{"article_info":{"article_id":110987,"cat_id":941,"cat_title":"Ôn tập hè Toán lớp 6","publish_time":1655887800,"cat_id2":941,"is_link":0,"thumbnail":"","title":"Dạng 5. Tìm x Chủ đề 6 Ôn hè Toán 6","label":"Dạng 5. Tìm x","alias":"dang-5-tim-x-chu-de-6-on-he-toan-6","introtext":"Dùng quy tắc thực hiện phép tính, quy tắc chuyển vế, quy tắc dấu ngoặc để đưa về các dạng quen thuộc để tìm x:","content":"<p><strong class=\"sub_question_head\" id=\"sub_question_head_1\">Lý thuyết</strong></p><p>Dùng quy tắc thực hiện phép tính, quy tắc chuyển vế, quy tắc dấu ngoặc để đưa về các dạng quen thuộc để tìm x:</p>\r\n<p>\\(\\begin{array}{l}1)x + a = b \\Rightarrow x = b - a\\\\2)x - a = b \\Rightarrow x = b + a\\\\3)a - x = b \\Rightarrow x = a - b\\\\4)a.x = b \\Rightarrow x = \\dfrac{b}{a}\\\\5)a:x = b \\Rightarrow x = \\dfrac{a}{b}\\\\6)x:a = b \\Rightarrow x = a.b\\\\7)\\dfrac{a}{b} = \\dfrac{x}{c} \\Rightarrow x = \\dfrac{{a.c}}{b}\\\\8){x^2} = {a^2} \\Rightarrow \\left[ {\\begin{array}{*{20}{c}}{x = a}\\\\{x =  - a}\\end{array}} \\right.\\\\9){x^3} = {a^3} \\Rightarrow x = a\\end{array}\\) </p><hr/><p><strong class=\"sub_question_head\" id=\"sub_question_head_2\">Bài tập</strong></p><p><strong>Bài 1:</strong></p>\r\n<p>Tìm x, biết:</p>\r\n<p>\\(\\begin{array}{l}a)2x + 3 = 1\\dfrac{2}{3}\\\\b)0,15 - 3x = {( - 10)^0}\\\\c) - x:\\dfrac{2}{5} = 0,8\\\\d)\\dfrac{{3x + 2}}{3} = \\dfrac{{ - 4}}{5}\\\\e)\\dfrac{{3x + 2}}{{ - 8}} = \\dfrac{{ - 2}}{{3x + 2}}\\\\f)\\left( {x + 1} \\right).\\left( { - 2x - 3} \\right) = 0\\end{array}\\)</p>\r\n<p><strong>Bài 2: </strong></p>\r\n<p>Tìm \\(x\\), biết:</p>\r\n<p>a) \\(\\dfrac{1}{3}x + \\dfrac{2}{5}\\left( {x - 1} \\right) = 0\\)                                                                   </p>\r\n<p>b) \\(3 \\cdot {\\left( {3x - \\dfrac{1}{2}} \\right)^3} + \\dfrac{1}{9} = 0\\)</p>\r\n<p>c) \\(3 \\cdot \\left( {1 - \\dfrac{1}{2}} \\right) - 5 \\cdot \\left( {x + \\dfrac{3}{5}} \\right) = {\\rm{ \\;}} - x + \\dfrac{1}{5}\\)                                     </p>\r\n<p>d) \\(\\dfrac{{3 - x}}{{5 - x}} = {\\left( {\\dfrac{{ - 3}}{5}} \\right)^2}\\)</p>\r\n<p>e) \\(x\\;:\\;\\dfrac{5}{8} = \\dfrac{{ - 13}}{{35}} \\cdot \\dfrac{{15}}{{ - 39}}\\)                                                           </p>\r\n<p>f) \\(\\left( {\\dfrac{7}{5}\\; + \\;x} \\right):\\dfrac{{25}}{{16}} = \\dfrac{{ - 4}}{5}\\)</p>\r\n<p>g) \\( - 4:\\left( {x + \\dfrac{{ - 2}}{3}} \\right) = \\dfrac{3}{4}\\)                                                                </p>\r\n<p>h) \\(\\left( {\\dfrac{{ - 1}}{5} + 2} \\right):\\left( {x - \\dfrac{7}{{10}}} \\right) = \\dfrac{{ - 1}}{4}\\)</p>\r\n<p><strong>Bài 3:</strong></p>\r\n<p>Tìm tập hợp các số nguyên <em>x</em> để: \\(\\dfrac{5}{6} + \\dfrac{{ - 7}}{8} \\le \\dfrac{x}{{24}} \\le \\dfrac{{ - 5}}{{12}} + \\dfrac{5}{8}\\)</p><p><strong>Lời giải chi tiết:</strong></p><p><strong>Bài 1:</strong></p>\r\n<p>Tìm x, biết:</p>\r\n<p>\\(\\begin{array}{l}a)2x + 3 = 1\\dfrac{2}{3}\\\\b)0,15 - 3x = {( - 10)^0}\\\\c) - x:\\dfrac{2}{5} = 0,8\\\\d)\\dfrac{{3x + 2}}{3} = \\dfrac{{ - 4}}{5}\\\\e)\\dfrac{{3x + 2}}{{ - 8}} = \\dfrac{{ - 2}}{{3x + 2}}\\\\f)\\left( {x + 1} \\right).\\left( { - 2x - 3} \\right) = 0\\end{array}\\)</p>\r\n<p><strong>Phương pháp</strong></p>\r\n<p>Áp dụng quy tắc thực hiện phép tính, quy tắc chuyển vế, quy tắc dấu ngoặc để đưa về các dạng quen thuộc để tìm x.</p>\r\n<p><strong>Lời giải</strong></p>\r\n<p>\\(\\begin{array}{l}a)2x + 3 = 1\\dfrac{2}{3}\\\\2x + 3 = \\dfrac{5}{3}\\\\2x = \\dfrac{5}{3} - 3\\\\2x = \\dfrac{5}{3} - \\dfrac{9}{3}\\\\2x = \\dfrac{{ - 4}}{3}\\\\x = \\dfrac{{ - 4}}{3}:2\\\\x = \\dfrac{{ - 4}}{3}.\\dfrac{1}{2}\\\\x = \\dfrac{{ - 2}}{3}\\end{array}\\)</p>\r\n<p>Vậy \\(x = \\dfrac{{ - 2}}{3}\\)</p>\r\n<p>\\(\\begin{array}{l}b)0,15 - 3x = {( - 10)^0}\\\\0,15 - 3x = 1\\\\3x = 0,15 - 1\\\\3x = 0,85\\\\3x = \\dfrac{{17}}{{20}}\\\\x = \\dfrac{{17}}{{20}}:3\\\\x = \\dfrac{{17}}{{20}}.\\dfrac{1}{3}\\\\x = \\dfrac{{17}}{{60}}\\end{array}\\)</p>\r\n<p>Vậy \\(x = \\dfrac{{17}}{{60}}\\)</p>\r\n<p>\\(\\begin{array}{l}c) - x:\\dfrac{2}{5} = 0,8\\\\ - x:0.4 = 0,8\\\\ - x = 0,8.0,4\\\\ - x = 0,32\\\\x =  - 0,32\\end{array}\\)</p>\r\n<p>Vậy x = -0,32</p>\r\n<p>\\(\\begin{array}{l}d)\\dfrac{{3x + 2}}{3} = \\dfrac{{ - 4}}{5}\\\\5.(3x + 2) = 3.( - 4)\\\\15x + 10 =  - 12\\\\15x =  - 12 - 10\\\\15x =  - 22\\\\x = \\dfrac{{ - 22}}{{15}}\\end{array}\\)</p>\r\n<p>Vậy \\(x = \\dfrac{{ - 22}}{{15}}\\)</p>\r\n<p>\\(\\begin{array}{l}e)\\dfrac{{3x + 2}}{{ - 8}} = \\dfrac{{ - 2}}{{3x + 2}}\\\\\\left( {3x + 2} \\right).\\left( {3x + 2} \\right) = ( - 8).( - 2)\\\\{\\left( {3x + 2} \\right)^2} = 16\\\\{\\left( {3x + 2} \\right)^2} = {4^2}\\\\\\left[ {\\begin{array}{*{20}{c}}{3x + 2 = 4}\\\\{3x + 2 =  - 4}\\end{array}} \\right.\\\\\\left[ {\\begin{array}{*{20}{c}}{3x = 2}\\\\{3x =  - 6}\\end{array}} \\right.\\\\\\left[ {\\begin{array}{*{20}{c}}{x = \\dfrac{2}{3}}\\\\{x =  - 2}\\end{array}} \\right.\\end{array}\\)</p>\r\n<p>Vậy \\(x \\in \\left\\{ {\\dfrac{2}{3}; - 2} \\right\\}\\)</p>\r\n<p>\\(\\begin{array}{l}f)\\left( {x + 1} \\right).\\left( { - 2x - 3} \\right) = 0\\\\\\left[ {\\begin{array}{*{20}{c}}{x + 1 = 0}\\\\{ - 2x - 3 = 0}\\end{array}} \\right.\\\\\\left[ {\\begin{array}{*{20}{c}}{x =  - 1}\\\\{x = \\dfrac{{ - 3}}{2}}\\end{array}} \\right.\\end{array}\\)</p>\r\n<p>Vậy \\(x \\in \\left\\{ { - 1;\\dfrac{{ - 3}}{2}} \\right\\}\\)</p>\r\n<p><strong>Bài 2: </strong></p>\r\n<p>Tìm \\(x\\), biết:</p>\r\n<p>a) \\(\\dfrac{1}{3}x + \\dfrac{2}{5}\\left( {x - 1} \\right) = 0\\)                                                                   </p>\r\n<p>b) \\(3 \\cdot {\\left( {3x - \\dfrac{1}{2}} \\right)^3} + \\dfrac{1}{9} = 0\\)</p>\r\n<p>c) \\(3 \\cdot \\left( {1 - \\dfrac{1}{2}} \\right) - 5 \\cdot \\left( {x + \\dfrac{3}{5}} \\right) = {\\rm{ \\;}} - x + \\dfrac{1}{5}\\)                                     </p>\r\n<p>d) \\(\\dfrac{{3 - x}}{{5 - x}} = {\\left( {\\dfrac{{ - 3}}{5}} \\right)^2}\\)</p>\r\n<p>e) \\(x\\;:\\;\\dfrac{5}{8} = \\dfrac{{ - 13}}{{35}} \\cdot \\dfrac{{15}}{{ - 39}}\\)                                                           </p>\r\n<p>f) \\(\\left( {\\dfrac{7}{5}\\; + \\;x} \\right):\\dfrac{{25}}{{16}} = \\dfrac{{ - 4}}{5}\\)</p>\r\n<p>g) \\( - 4:\\left( {x + \\dfrac{{ - 2}}{3}} \\right) = \\dfrac{3}{4}\\)                                                                </p>\r\n<p>h) \\(\\left( {\\dfrac{{ - 1}}{5} + 2} \\right):\\left( {x - \\dfrac{7}{{10}}} \\right) = \\dfrac{{ - 1}}{4}\\)</p>\r\n<p><strong>Phương pháp</strong></p>\r\n<p>Áp dụng các qui tắc cộng, trừ, nhân, chia phân số, qui tắc tính giá trị của biểu thức.</p>\r\n<p><strong>Lời giải</strong></p>\r\n<table style=\"width: 718px;\" border=\"0\" cellspacing=\"0\" cellpadding=\"0\">\r\n<tbody>\r\n<tr>\r\n<td colspan=\"2\" valign=\"top\" width=\"359\">\r\n<p>a) \\(\\dfrac{1}{3}x + \\dfrac{2}{5}\\left( {x - 1} \\right) = 0\\)</p>\r\n<p>    \\(\\begin{array}{l}\\dfrac{1}{3}x + \\dfrac{2}{5}x - \\dfrac{2}{5} = 0\\\\\\left( {\\dfrac{1}{3} + \\dfrac{2}{5}} \\right)x = \\dfrac{2}{5}\\\\\\dfrac{{11}}{{15}}x = \\dfrac{2}{5}\\end{array}\\)</p>\r\n<p>          \\(x = \\dfrac{2}{5}:\\dfrac{{11}}{{15}}\\)</p>\r\n<p>       \\(\\begin{array}{l}x = \\dfrac{2}{5} \\cdot \\dfrac{{15}}{{11}}\\\\x = \\dfrac{6}{{11}}\\end{array}\\)</p>\r\n<p>Vậy \\(x = \\dfrac{6}{{11}} \\cdot \\)</p>\r\n<p><strong> </strong>b) \\(3.{\\left( {3x - \\dfrac{1}{2}} \\right)^3} + \\dfrac{1}{9} = 0\\)</p>\r\n<p>    \\(\\begin{array}{l}3.{\\left( {3x - \\dfrac{1}{2}} \\right)^3} =  - \\dfrac{1}{9}\\\\{\\left( {3x - \\dfrac{1}{2}} \\right)^3} =  - \\dfrac{1}{9}:3\\\\{\\left( {3x - \\dfrac{1}{2}} \\right)^3} =  - \\dfrac{1}{{27}} = \\left( {\\dfrac{{ - 1}}{3}} \\right)\\end{array}\\)</p>\r\n<p>\\( \\Rightarrow 3x - \\dfrac{1}{2} = {\\dfrac{{ - 1}}{3}^3}\\)</p>\r\n<p>\\(\\begin{array}{l}{\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} 3x = \\dfrac{{ - 1}}{3} + \\dfrac{1}{2}{\\kern 1pt} {\\kern 1pt} \\\\{\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} 3x = \\dfrac{{ - 2}}{6} + \\dfrac{3}{6}\\\\{\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} 3x = \\dfrac{1}{6}{\\kern 1pt} {\\kern 1pt} \\\\{\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} x = \\dfrac{1}{{18}}\\end{array}\\)</p>\r\n<p>Vậy \\(x = \\dfrac{1}{{18}} \\cdot \\)</p>\r\n<p><strong> </strong></p>\r\n</td>\r\n<td valign=\"top\" width=\"339\">\r\n<p> </p>\r\n</td>\r\n<td width=\"20\">\r\n<p> </p>\r\n</td>\r\n</tr>\r\n<tr>\r\n<td colspan=\"2\" valign=\"top\" width=\"359\">\r\n<p>c) \\(3.\\left( {1 - \\dfrac{1}{2}} \\right) - 5\\left( {x + \\dfrac{3}{5}} \\right) = {\\rm{ \\;}} - x + \\dfrac{1}{5}\\)</p>\r\n<p>    \\(\\begin{array}{*{20}{l}}{3 - \\dfrac{3}{2} - \\left( {5x + 5.\\dfrac{3}{5}} \\right) = {\\rm{ \\;}} - x + \\dfrac{1}{5}}\\\\{\\dfrac{3}{2} - 5x - 3 = {\\rm{ \\;}} - x + \\dfrac{1}{5}}\\\\{ - 5x + x = \\dfrac{1}{5} - \\dfrac{3}{2} + 3}\\end{array}\\)</p>\r\n<p> \\(\\begin{array}{*{20}{l}}{ - 4x = \\dfrac{{ - 13}}{{10}} + 3}\\\\{ - 4x = \\dfrac{{17}}{{10}}}\\\\{x = \\dfrac{{17}}{{10}}:\\left( { - 4} \\right)}\\\\{x = {\\rm{ \\;}} - \\dfrac{{17}}{{40}}}\\end{array}\\)</p>\r\n<p>Vậy \\(x = {\\rm{ \\;}} - \\dfrac{{17}}{{40}} \\cdot \\)</p>\r\n<p>d) \\(\\dfrac{{3 - x}}{{5 - x}} = {\\left( {\\dfrac{{ - 3}}{5}} \\right)^2}\\)</p>\r\n<p>Điều kiện: \\(5 - x \\ne 0 \\Leftrightarrow x \\ne 5.\\)</p>\r\n<p>\\(\\begin{array}{*{20}{l}}{ \\Rightarrow \\dfrac{{3 - x}}{{5 - x}} = \\dfrac{9}{{25}}}\\\\{ \\Rightarrow \\left( {3 - x} \\right).25 = 9.\\left( {5 - x} \\right)}\\\\{{\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} 75 - 25x = 45 - 9x{\\kern 1pt} }\\\\{{\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt}  - 25x + 9x = 45 - 75}\\\\{{\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt}  - 16x = {\\rm{ \\;}} - 30}\\\\{{\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} x = \\dfrac{{ - 30}}{{ - 16}} = \\dfrac{{15}}{8}}\\end{array}\\)</p>\r\n<p>Vậy \\(x = \\dfrac{{15}}{8} \\cdot \\)</p>\r\n<p> </p>\r\n<p><strong> </strong></p>\r\n</td>\r\n<td valign=\"top\" width=\"339\">\r\n<p> </p>\r\n</td>\r\n<td width=\"20\">\r\n<p> </p>\r\n</td>\r\n</tr>\r\n<tr>\r\n<td valign=\"top\" width=\"359\">\r\n<p>e) \\(x\\;:\\;\\dfrac{5}{8} = \\dfrac{{ - 13}}{{35}} \\cdot \\dfrac{{15}}{{ - 39}}\\)</p>\r\n<p>     \\(\\begin{array}{*{20}{l}}{x:\\dfrac{5}{8} = \\dfrac{1}{7}}\\\\{x\\;\\;\\;\\;\\; = \\dfrac{1}{7} \\cdot \\dfrac{5}{8}}\\\\{x\\;\\;\\;\\;\\; = \\dfrac{5}{{56}}.}\\end{array}\\)</p>\r\n<p>Vậy \\(x = \\dfrac{5}{{56}}\\)</p>\r\n<p> f) \\(\\left( {\\dfrac{7}{5}\\; + \\;x} \\right):\\dfrac{{25}}{{16}} = \\dfrac{{ - 4}}{5}\\)</p>\r\n<p>     \\(\\begin{array}{*{20}{l}}{\\dfrac{7}{5}\\; + \\;x\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; = \\dfrac{{ - 4}}{5} \\cdot \\dfrac{{25}}{{16}}}\\\\{\\dfrac{7}{5}\\; + \\;x\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; = \\dfrac{{ - 5}}{4}}\\\\{\\;\\;{\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} x\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; = \\dfrac{{ - 5}}{4} - \\dfrac{7}{5}}\\\\{\\;\\;\\;\\;\\;\\;x\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; = \\dfrac{{ - 53}}{{20}}.}\\end{array}\\)</p>\r\n<p>Vậy \\(x = \\dfrac{{ - 53}}{{20}}\\).</p>\r\n</td>\r\n<td colspan=\"3\" valign=\"top\" width=\"359\">\r\n<p> </p>\r\n</td>\r\n</tr>\r\n<tr>\r\n<td valign=\"top\" width=\"359\">\r\n<p>g) \\( - 4:\\left( {x + \\dfrac{{ - 2}}{3}} \\right) = \\dfrac{3}{4}\\)</p>\r\n<p>\\(\\begin{array}{*{20}{l}}{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;x + \\dfrac{{ - 2}}{3} = {\\rm{ \\;}} - 4:\\dfrac{3}{4}}\\\\{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;x + \\dfrac{{ - 2}}{3} = {\\rm{ \\;}} - 4.\\dfrac{4}{3}}\\\\{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;x + \\dfrac{{ - 2}}{3} = \\dfrac{{ - 16}}{3}}\\\\{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;x\\;\\;\\;\\;\\;\\;\\;\\; = \\dfrac{{ - 16}}{3} - \\left( {\\dfrac{{ - 2}}{3}} \\right)}\\\\{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;x\\;\\;\\;\\;\\;\\;\\;\\; = \\dfrac{{ - 14}}{3}.}\\end{array}\\)</p>\r\n<p>Vậy \\(x = \\dfrac{{ - 14}}{3}\\).</p>\r\n<p> h) \\(\\left( {\\dfrac{{ - 1}}{5} + 2} \\right):\\left( {x - \\dfrac{7}{{10}}} \\right) = \\dfrac{{ - 1}}{4}\\)</p>\r\n<p>\\(\\dfrac{{ - 1 + 10}}{5}:\\left( {x - \\dfrac{7}{{10}}} \\right) = \\dfrac{{ - 1}}{4}\\)</p>\r\n<p>\\(\\begin{array}{*{20}{l}}{\\;\\;\\;\\;\\dfrac{9}{5}:\\left( {x - \\dfrac{7}{{10}}} \\right) = \\dfrac{{ - 1}}{4}}\\\\{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;x - \\dfrac{7}{{10}} = \\dfrac{9}{5}:\\dfrac{{ - 1}}{4}}\\\\{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;x - \\dfrac{7}{{10}} = \\dfrac{9}{5}.\\left( { - 4} \\right)}\\\\{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;x - \\dfrac{7}{{10}} = \\dfrac{{ - 36}}{5}}\\\\{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;x\\;\\;\\;\\;\\;\\;\\;\\; = \\dfrac{{ - 36}}{5} + \\dfrac{7}{{10}}}\\\\{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;x\\;\\;\\;\\;\\;\\;\\;\\; = \\dfrac{{ - 13}}{2}.}\\end{array}\\)</p>\r\n<p>      Vậy \\(x = \\dfrac{{ - 13}}{2}\\)</p>\r\n<p> </p>\r\n</td>\r\n<td colspan=\"3\" valign=\"top\" width=\"359\">\r\n<p> </p>\r\n<p> </p>\r\n</td>\r\n</tr>\r\n<tr>\r\n<td width=\"348\"> </td>\r\n<td width=\"0\"> </td>\r\n<td width=\"392\"> </td>\r\n<td width=\"13\"> </td>\r\n</tr>\r\n</tbody>\r\n</table>\r\n<p> </p>\r\n<p><strong>Bài 3:</strong></p>\r\n<p>Tìm tập hợp các số nguyên <em>x</em> để: \\(\\dfrac{5}{6} + \\dfrac{{ - 7}}{8} \\le \\dfrac{x}{{24}} \\le \\dfrac{{ - 5}}{{12}} + \\dfrac{5}{8}\\)</p>\r\n<p><strong>Phương pháp</strong></p>\r\n<p>+ Thực hiện phép cộng các phân số đã biết.</p>\r\n<p>+ Xác định vai trò của số chưa biết trong phép toán rồi kết luận.</p>\r\n<p><strong>Lời giải</strong></p>\r\n<p>\\(\\begin{array}{*{20}{l}}{\\dfrac{5}{6} + \\dfrac{{ - 7}}{8} \\le \\dfrac{x}{{24}} \\le \\dfrac{{ - 5}}{{12}} + \\dfrac{5}{8}}\\\\{ \\Rightarrow \\dfrac{{ - 1}}{{24}} \\le \\dfrac{x}{{24}} \\le \\dfrac{5}{{24}}}\\\\{ \\Rightarrow {\\rm{ \\;}} - 1 \\le x \\le 5}\\end{array}\\)</p>\r\n<p>Vì \\(x \\in \\mathbb{Z}\\) nên \\(x \\in \\left\\{ { - 1;0;1;2;3;4;5} \\right\\}\\)</p>\r\n<p>Vậy \\(x \\in \\left\\{ { - 1;0;1;2;3;4;5} \\right\\}\\) </p>","common_article":"","display_type":2,"article_type":1,"cmt_count":6,"zip_status":1,"last_modified_time":1655888355,"last_zip_time":1655888355,"zip_link":"https://offline.loigiaihay.com/articles/941/110987/110987_1655888355.zip","picture":"","youtube_id":"","url_upload_pdf":"https://img.loigiaihay.com/picture/share_img/2022/0622/dang-5-tim-x-chu-de-6-on-he-toan-6-1655888398.pdf","video_start_time":"0","quizlet_id":"0","ref_article_id":0,"readmore_display_type":0,"wiki":0,"sub_articles":[{"id":"440030","group_title":"","sub_article_link":"0","order_in_article":"1","title":"Lý thuyết","content":"<p>Dùng quy tắc thực hiện phép tính, quy tắc chuyển vế, quy tắc dấu ngoặc để đưa về các dạng quen thuộc để tìm x:</p>\r\n<p>\\(\\begin{array}{l}1)x + a = b \\Rightarrow x = b - a\\\\2)x - a = b \\Rightarrow x = b + a\\\\3)a - x = b \\Rightarrow x = a - b\\\\4)a.x = b \\Rightarrow x = \\dfrac{b}{a}\\\\5)a:x = b \\Rightarrow x = \\dfrac{a}{b}\\\\6)x:a = b \\Rightarrow x = a.b\\\\7)\\dfrac{a}{b} = \\dfrac{x}{c} \\Rightarrow x = \\dfrac{{a.c}}{b}\\\\8){x^2} = {a^2} \\Rightarrow \\left[ {\\begin{array}{*{20}{c}}{x = a}\\\\{x =  - a}\\end{array}} \\right.\\\\9){x^3} = {a^3} \\Rightarrow x = a\\end{array}\\) </p>","youtube_id":"","video_start_time":"0"},{"id":"440031","group_title":"","sub_article_link":"0","order_in_article":"2","title":"Bài tập","content":"<p><strong>Bài 1:</strong></p>\r\n<p>Tìm x, biết:</p>\r\n<p>\\(\\begin{array}{l}a)2x + 3 = 1\\dfrac{2}{3}\\\\b)0,15 - 3x = {( - 10)^0}\\\\c) - x:\\dfrac{2}{5} = 0,8\\\\d)\\dfrac{{3x + 2}}{3} = \\dfrac{{ - 4}}{5}\\\\e)\\dfrac{{3x + 2}}{{ - 8}} = \\dfrac{{ - 2}}{{3x + 2}}\\\\f)\\left( {x + 1} \\right).\\left( { - 2x - 3} \\right) = 0\\end{array}\\)</p>\r\n<p><strong>Bài 2: </strong></p>\r\n<p>Tìm \\(x\\), biết:</p>\r\n<p>a) \\(\\dfrac{1}{3}x + \\dfrac{2}{5}\\left( {x - 1} \\right) = 0\\)                                                                   </p>\r\n<p>b) \\(3 \\cdot {\\left( {3x - \\dfrac{1}{2}} \\right)^3} + \\dfrac{1}{9} = 0\\)</p>\r\n<p>c) \\(3 \\cdot \\left( {1 - \\dfrac{1}{2}} \\right) - 5 \\cdot \\left( {x + \\dfrac{3}{5}} \\right) = {\\rm{ \\;}} - x + \\dfrac{1}{5}\\)                                     </p>\r\n<p>d) \\(\\dfrac{{3 - x}}{{5 - x}} = {\\left( {\\dfrac{{ - 3}}{5}} \\right)^2}\\)</p>\r\n<p>e) \\(x\\;:\\;\\dfrac{5}{8} = \\dfrac{{ - 13}}{{35}} \\cdot \\dfrac{{15}}{{ - 39}}\\)                                                           </p>\r\n<p>f) \\(\\left( {\\dfrac{7}{5}\\; + \\;x} \\right):\\dfrac{{25}}{{16}} = \\dfrac{{ - 4}}{5}\\)</p>\r\n<p>g) \\( - 4:\\left( {x + \\dfrac{{ - 2}}{3}} \\right) = \\dfrac{3}{4}\\)                                                                </p>\r\n<p>h) \\(\\left( {\\dfrac{{ - 1}}{5} + 2} \\right):\\left( {x - \\dfrac{7}{{10}}} \\right) = \\dfrac{{ - 1}}{4}\\)</p>\r\n<p><strong>Bài 3:</strong></p>\r\n<p>Tìm tập hợp các số nguyên <em>x</em> để: \\(\\dfrac{5}{6} + \\dfrac{{ - 7}}{8} \\le \\dfrac{x}{{24}} \\le \\dfrac{{ - 5}}{{12}} + \\dfrac{5}{8}\\)</p><p><strong>Lời giải chi tiết:</strong></p><p><strong>Bài 1:</strong></p>\r\n<p>Tìm x, biết:</p>\r\n<p>\\(\\begin{array}{l}a)2x + 3 = 1\\dfrac{2}{3}\\\\b)0,15 - 3x = {( - 10)^0}\\\\c) - x:\\dfrac{2}{5} = 0,8\\\\d)\\dfrac{{3x + 2}}{3} = \\dfrac{{ - 4}}{5}\\\\e)\\dfrac{{3x + 2}}{{ - 8}} = \\dfrac{{ - 2}}{{3x + 2}}\\\\f)\\left( {x + 1} \\right).\\left( { - 2x - 3} \\right) = 0\\end{array}\\)</p>\r\n<p><strong>Phương pháp</strong></p>\r\n<p>Áp dụng quy tắc thực hiện phép tính, quy tắc chuyển vế, quy tắc dấu ngoặc để đưa về các dạng quen thuộc để tìm x.</p>\r\n<p><strong>Lời giải</strong></p>\r\n<p>\\(\\begin{array}{l}a)2x + 3 = 1\\dfrac{2}{3}\\\\2x + 3 = \\dfrac{5}{3}\\\\2x = \\dfrac{5}{3} - 3\\\\2x = \\dfrac{5}{3} - \\dfrac{9}{3}\\\\2x = \\dfrac{{ - 4}}{3}\\\\x = \\dfrac{{ - 4}}{3}:2\\\\x = \\dfrac{{ - 4}}{3}.\\dfrac{1}{2}\\\\x = \\dfrac{{ - 2}}{3}\\end{array}\\)</p>\r\n<p>Vậy \\(x = \\dfrac{{ - 2}}{3}\\)</p>\r\n<p>\\(\\begin{array}{l}b)0,15 - 3x = {( - 10)^0}\\\\0,15 - 3x = 1\\\\3x = 0,15 - 1\\\\3x = 0,85\\\\3x = \\dfrac{{17}}{{20}}\\\\x = \\dfrac{{17}}{{20}}:3\\\\x = \\dfrac{{17}}{{20}}.\\dfrac{1}{3}\\\\x = \\dfrac{{17}}{{60}}\\end{array}\\)</p>\r\n<p>Vậy \\(x = \\dfrac{{17}}{{60}}\\)</p>\r\n<p>\\(\\begin{array}{l}c) - x:\\dfrac{2}{5} = 0,8\\\\ - x:0.4 = 0,8\\\\ - x = 0,8.0,4\\\\ - x = 0,32\\\\x =  - 0,32\\end{array}\\)</p>\r\n<p>Vậy x = -0,32</p>\r\n<p>\\(\\begin{array}{l}d)\\dfrac{{3x + 2}}{3} = \\dfrac{{ - 4}}{5}\\\\5.(3x + 2) = 3.( - 4)\\\\15x + 10 =  - 12\\\\15x =  - 12 - 10\\\\15x =  - 22\\\\x = \\dfrac{{ - 22}}{{15}}\\end{array}\\)</p>\r\n<p>Vậy \\(x = \\dfrac{{ - 22}}{{15}}\\)</p>\r\n<p>\\(\\begin{array}{l}e)\\dfrac{{3x + 2}}{{ - 8}} = \\dfrac{{ - 2}}{{3x + 2}}\\\\\\left( {3x + 2} \\right).\\left( {3x + 2} \\right) = ( - 8).( - 2)\\\\{\\left( {3x + 2} \\right)^2} = 16\\\\{\\left( {3x + 2} \\right)^2} = {4^2}\\\\\\left[ {\\begin{array}{*{20}{c}}{3x + 2 = 4}\\\\{3x + 2 =  - 4}\\end{array}} \\right.\\\\\\left[ {\\begin{array}{*{20}{c}}{3x = 2}\\\\{3x =  - 6}\\end{array}} \\right.\\\\\\left[ {\\begin{array}{*{20}{c}}{x = \\dfrac{2}{3}}\\\\{x =  - 2}\\end{array}} \\right.\\end{array}\\)</p>\r\n<p>Vậy \\(x \\in \\left\\{ {\\dfrac{2}{3}; - 2} \\right\\}\\)</p>\r\n<p>\\(\\begin{array}{l}f)\\left( {x + 1} \\right).\\left( { - 2x - 3} \\right) = 0\\\\\\left[ {\\begin{array}{*{20}{c}}{x + 1 = 0}\\\\{ - 2x - 3 = 0}\\end{array}} \\right.\\\\\\left[ {\\begin{array}{*{20}{c}}{x =  - 1}\\\\{x = \\dfrac{{ - 3}}{2}}\\end{array}} \\right.\\end{array}\\)</p>\r\n<p>Vậy \\(x \\in \\left\\{ { - 1;\\dfrac{{ - 3}}{2}} \\right\\}\\)</p>\r\n<p><strong>Bài 2: </strong></p>\r\n<p>Tìm \\(x\\), biết:</p>\r\n<p>a) \\(\\dfrac{1}{3}x + \\dfrac{2}{5}\\left( {x - 1} \\right) = 0\\)                                                                   </p>\r\n<p>b) \\(3 \\cdot {\\left( {3x - \\dfrac{1}{2}} \\right)^3} + \\dfrac{1}{9} = 0\\)</p>\r\n<p>c) \\(3 \\cdot \\left( {1 - \\dfrac{1}{2}} \\right) - 5 \\cdot \\left( {x + \\dfrac{3}{5}} \\right) = {\\rm{ \\;}} - x + \\dfrac{1}{5}\\)                                     </p>\r\n<p>d) \\(\\dfrac{{3 - x}}{{5 - x}} = {\\left( {\\dfrac{{ - 3}}{5}} \\right)^2}\\)</p>\r\n<p>e) \\(x\\;:\\;\\dfrac{5}{8} = \\dfrac{{ - 13}}{{35}} \\cdot \\dfrac{{15}}{{ - 39}}\\)                                                           </p>\r\n<p>f) \\(\\left( {\\dfrac{7}{5}\\; + \\;x} \\right):\\dfrac{{25}}{{16}} = \\dfrac{{ - 4}}{5}\\)</p>\r\n<p>g) \\( - 4:\\left( {x + \\dfrac{{ - 2}}{3}} \\right) = \\dfrac{3}{4}\\)                                                                </p>\r\n<p>h) \\(\\left( {\\dfrac{{ - 1}}{5} + 2} \\right):\\left( {x - \\dfrac{7}{{10}}} \\right) = \\dfrac{{ - 1}}{4}\\)</p>\r\n<p><strong>Phương pháp</strong></p>\r\n<p>Áp dụng các qui tắc cộng, trừ, nhân, chia phân số, qui tắc tính giá trị của biểu thức.</p>\r\n<p><strong>Lời giải</strong></p>\r\n<table style=\"width: 718px;\" border=\"0\" cellspacing=\"0\" cellpadding=\"0\">\r\n<tbody>\r\n<tr>\r\n<td colspan=\"2\" valign=\"top\" width=\"359\">\r\n<p>a) \\(\\dfrac{1}{3}x + \\dfrac{2}{5}\\left( {x - 1} \\right) = 0\\)</p>\r\n<p>    \\(\\begin{array}{l}\\dfrac{1}{3}x + \\dfrac{2}{5}x - \\dfrac{2}{5} = 0\\\\\\left( {\\dfrac{1}{3} + \\dfrac{2}{5}} \\right)x = \\dfrac{2}{5}\\\\\\dfrac{{11}}{{15}}x = \\dfrac{2}{5}\\end{array}\\)</p>\r\n<p>          \\(x = \\dfrac{2}{5}:\\dfrac{{11}}{{15}}\\)</p>\r\n<p>       \\(\\begin{array}{l}x = \\dfrac{2}{5} \\cdot \\dfrac{{15}}{{11}}\\\\x = \\dfrac{6}{{11}}\\end{array}\\)</p>\r\n<p>Vậy \\(x = \\dfrac{6}{{11}} \\cdot \\)</p>\r\n<p><strong> </strong>b) \\(3.{\\left( {3x - \\dfrac{1}{2}} \\right)^3} + \\dfrac{1}{9} = 0\\)</p>\r\n<p>    \\(\\begin{array}{l}3.{\\left( {3x - \\dfrac{1}{2}} \\right)^3} =  - \\dfrac{1}{9}\\\\{\\left( {3x - \\dfrac{1}{2}} \\right)^3} =  - \\dfrac{1}{9}:3\\\\{\\left( {3x - \\dfrac{1}{2}} \\right)^3} =  - \\dfrac{1}{{27}} = \\left( {\\dfrac{{ - 1}}{3}} \\right)\\end{array}\\)</p>\r\n<p>\\( \\Rightarrow 3x - \\dfrac{1}{2} = {\\dfrac{{ - 1}}{3}^3}\\)</p>\r\n<p>\\(\\begin{array}{l}{\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} 3x = \\dfrac{{ - 1}}{3} + \\dfrac{1}{2}{\\kern 1pt} {\\kern 1pt} \\\\{\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} 3x = \\dfrac{{ - 2}}{6} + \\dfrac{3}{6}\\\\{\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} 3x = \\dfrac{1}{6}{\\kern 1pt} {\\kern 1pt} \\\\{\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} x = \\dfrac{1}{{18}}\\end{array}\\)</p>\r\n<p>Vậy \\(x = \\dfrac{1}{{18}} \\cdot \\)</p>\r\n<p><strong> </strong></p>\r\n</td>\r\n<td valign=\"top\" width=\"339\">\r\n<p> </p>\r\n</td>\r\n<td width=\"20\">\r\n<p> </p>\r\n</td>\r\n</tr>\r\n<tr>\r\n<td colspan=\"2\" valign=\"top\" width=\"359\">\r\n<p>c) \\(3.\\left( {1 - \\dfrac{1}{2}} \\right) - 5\\left( {x + \\dfrac{3}{5}} \\right) = {\\rm{ \\;}} - x + \\dfrac{1}{5}\\)</p>\r\n<p>    \\(\\begin{array}{*{20}{l}}{3 - \\dfrac{3}{2} - \\left( {5x + 5.\\dfrac{3}{5}} \\right) = {\\rm{ \\;}} - x + \\dfrac{1}{5}}\\\\{\\dfrac{3}{2} - 5x - 3 = {\\rm{ \\;}} - x + \\dfrac{1}{5}}\\\\{ - 5x + x = \\dfrac{1}{5} - \\dfrac{3}{2} + 3}\\end{array}\\)</p>\r\n<p> \\(\\begin{array}{*{20}{l}}{ - 4x = \\dfrac{{ - 13}}{{10}} + 3}\\\\{ - 4x = \\dfrac{{17}}{{10}}}\\\\{x = \\dfrac{{17}}{{10}}:\\left( { - 4} \\right)}\\\\{x = {\\rm{ \\;}} - \\dfrac{{17}}{{40}}}\\end{array}\\)</p>\r\n<p>Vậy \\(x = {\\rm{ \\;}} - \\dfrac{{17}}{{40}} \\cdot \\)</p>\r\n<p>d) \\(\\dfrac{{3 - x}}{{5 - x}} = {\\left( {\\dfrac{{ - 3}}{5}} \\right)^2}\\)</p>\r\n<p>Điều kiện: \\(5 - x \\ne 0 \\Leftrightarrow x \\ne 5.\\)</p>\r\n<p>\\(\\begin{array}{*{20}{l}}{ \\Rightarrow \\dfrac{{3 - x}}{{5 - x}} = \\dfrac{9}{{25}}}\\\\{ \\Rightarrow \\left( {3 - x} \\right).25 = 9.\\left( {5 - x} \\right)}\\\\{{\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} 75 - 25x = 45 - 9x{\\kern 1pt} }\\\\{{\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt}  - 25x + 9x = 45 - 75}\\\\{{\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt}  - 16x = {\\rm{ \\;}} - 30}\\\\{{\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} x = \\dfrac{{ - 30}}{{ - 16}} = \\dfrac{{15}}{8}}\\end{array}\\)</p>\r\n<p>Vậy \\(x = \\dfrac{{15}}{8} \\cdot \\)</p>\r\n<p> </p>\r\n<p><strong> </strong></p>\r\n</td>\r\n<td valign=\"top\" width=\"339\">\r\n<p> </p>\r\n</td>\r\n<td width=\"20\">\r\n<p> </p>\r\n</td>\r\n</tr>\r\n<tr>\r\n<td valign=\"top\" width=\"359\">\r\n<p>e) \\(x\\;:\\;\\dfrac{5}{8} = \\dfrac{{ - 13}}{{35}} \\cdot \\dfrac{{15}}{{ - 39}}\\)</p>\r\n<p>     \\(\\begin{array}{*{20}{l}}{x:\\dfrac{5}{8} = \\dfrac{1}{7}}\\\\{x\\;\\;\\;\\;\\; = \\dfrac{1}{7} \\cdot \\dfrac{5}{8}}\\\\{x\\;\\;\\;\\;\\; = \\dfrac{5}{{56}}.}\\end{array}\\)</p>\r\n<p>Vậy \\(x = \\dfrac{5}{{56}}\\)</p>\r\n<p> f) \\(\\left( {\\dfrac{7}{5}\\; + \\;x} \\right):\\dfrac{{25}}{{16}} = \\dfrac{{ - 4}}{5}\\)</p>\r\n<p>     \\(\\begin{array}{*{20}{l}}{\\dfrac{7}{5}\\; + \\;x\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; = \\dfrac{{ - 4}}{5} \\cdot \\dfrac{{25}}{{16}}}\\\\{\\dfrac{7}{5}\\; + \\;x\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; = \\dfrac{{ - 5}}{4}}\\\\{\\;\\;{\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} {\\kern 1pt} x\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; = \\dfrac{{ - 5}}{4} - \\dfrac{7}{5}}\\\\{\\;\\;\\;\\;\\;\\;x\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; = \\dfrac{{ - 53}}{{20}}.}\\end{array}\\)</p>\r\n<p>Vậy \\(x = \\dfrac{{ - 53}}{{20}}\\).</p>\r\n</td>\r\n<td colspan=\"3\" valign=\"top\" width=\"359\">\r\n<p> </p>\r\n</td>\r\n</tr>\r\n<tr>\r\n<td valign=\"top\" width=\"359\">\r\n<p>g) \\( - 4:\\left( {x + \\dfrac{{ - 2}}{3}} \\right) = \\dfrac{3}{4}\\)</p>\r\n<p>\\(\\begin{array}{*{20}{l}}{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;x + \\dfrac{{ - 2}}{3} = {\\rm{ \\;}} - 4:\\dfrac{3}{4}}\\\\{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;x + \\dfrac{{ - 2}}{3} = {\\rm{ \\;}} - 4.\\dfrac{4}{3}}\\\\{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;x + \\dfrac{{ - 2}}{3} = \\dfrac{{ - 16}}{3}}\\\\{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;x\\;\\;\\;\\;\\;\\;\\;\\; = \\dfrac{{ - 16}}{3} - \\left( {\\dfrac{{ - 2}}{3}} \\right)}\\\\{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;x\\;\\;\\;\\;\\;\\;\\;\\; = \\dfrac{{ - 14}}{3}.}\\end{array}\\)</p>\r\n<p>Vậy \\(x = \\dfrac{{ - 14}}{3}\\).</p>\r\n<p> h) \\(\\left( {\\dfrac{{ - 1}}{5} + 2} \\right):\\left( {x - \\dfrac{7}{{10}}} \\right) = \\dfrac{{ - 1}}{4}\\)</p>\r\n<p>\\(\\dfrac{{ - 1 + 10}}{5}:\\left( {x - \\dfrac{7}{{10}}} \\right) = \\dfrac{{ - 1}}{4}\\)</p>\r\n<p>\\(\\begin{array}{*{20}{l}}{\\;\\;\\;\\;\\dfrac{9}{5}:\\left( {x - \\dfrac{7}{{10}}} \\right) = \\dfrac{{ - 1}}{4}}\\\\{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;x - \\dfrac{7}{{10}} = \\dfrac{9}{5}:\\dfrac{{ - 1}}{4}}\\\\{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;x - \\dfrac{7}{{10}} = \\dfrac{9}{5}.\\left( { - 4} \\right)}\\\\{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;x - \\dfrac{7}{{10}} = \\dfrac{{ - 36}}{5}}\\\\{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;x\\;\\;\\;\\;\\;\\;\\;\\; = \\dfrac{{ - 36}}{5} + \\dfrac{7}{{10}}}\\\\{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;x\\;\\;\\;\\;\\;\\;\\;\\; = \\dfrac{{ - 13}}{2}.}\\end{array}\\)</p>\r\n<p>      Vậy \\(x = \\dfrac{{ - 13}}{2}\\)</p>\r\n<p> </p>\r\n</td>\r\n<td colspan=\"3\" valign=\"top\" width=\"359\">\r\n<p> </p>\r\n<p> </p>\r\n</td>\r\n</tr>\r\n<tr>\r\n<td width=\"348\"> </td>\r\n<td width=\"0\"> </td>\r\n<td width=\"392\"> </td>\r\n<td width=\"13\"> </td>\r\n</tr>\r\n</tbody>\r\n</table>\r\n<p> </p>\r\n<p><strong>Bài 3:</strong></p>\r\n<p>Tìm tập hợp các số nguyên <em>x</em> để: \\(\\dfrac{5}{6} + \\dfrac{{ - 7}}{8} \\le \\dfrac{x}{{24}} \\le \\dfrac{{ - 5}}{{12}} + \\dfrac{5}{8}\\)</p>\r\n<p><strong>Phương pháp</strong></p>\r\n<p>+ Thực hiện phép cộng các phân số đã biết.</p>\r\n<p>+ Xác định vai trò của số chưa biết trong phép toán rồi kết luận.</p>\r\n<p><strong>Lời giải</strong></p>\r\n<p>\\(\\begin{array}{*{20}{l}}{\\dfrac{5}{6} + \\dfrac{{ - 7}}{8} \\le \\dfrac{x}{{24}} \\le \\dfrac{{ - 5}}{{12}} + \\dfrac{5}{8}}\\\\{ \\Rightarrow \\dfrac{{ - 1}}{{24}} \\le \\dfrac{x}{{24}} \\le \\dfrac{5}{{24}}}\\\\{ \\Rightarrow {\\rm{ \\;}} - 1 \\le x \\le 5}\\end{array}\\)</p>\r\n<p>Vì \\(x \\in \\mathbb{Z}\\) nên \\(x \\in \\left\\{ { - 1;0;1;2;3;4;5} \\right\\}\\)</p>\r\n<p>Vậy \\(x \\in \\left\\{ { - 1;0;1;2;3;4;5} \\right\\}\\) </p>","youtube_id":"","video_start_time":"0"}],"comment_url":"https://loigiaihay.com/comment/dang-5-tim-x-chu-de-6-on-he-toan-6-c941a110987.html","origin_url":"https://loigiaihay.com/dang-5-tim-x-chu-de-6-on-he-toan-6-c941a110987.html","relate_link":[]},"callouts":[],"relate_articles":[],"others_in_cat":[{"article_id":110990,"cat_id":941,"publish_time":1655888520,"is_link":0,"thumbnail":"","title":"Dạng 6. Dãy phân số viết theo quy luật Chủ đề 6 Ôn hè Toán 6","alias":"dang-6-day-phan-so-viet-theo-quy-luat-chu-de-6-on-he-toan-6","introtext":"Phát hiện quy luật của dãy số","display_type":2,"article_type":1},{"article_id":111045,"cat_id":941,"publish_time":1655967180,"is_link":0,"thumbnail":"","title":"Dạng 7. Hai bài toán về phân số Chủ đề 6 Ôn hè Toán 6","alias":"dang-7-hai-bai-toan-ve-phan-so-chu-de-6-on-he-toan-6","introtext":"a) Tìm a/b của m là ","display_type":2,"article_type":1},{"article_id":110939,"cat_id":941,"publish_time":1655804340,"is_link":0,"thumbnail":"","title":"Dạng 4. Tính bằng cách hợp lí Chủ đề 6 Ôn hè Toán 6","alias":"dang-4-tinh-bang-cach-hop-li-chu-de-6-on-he-toan-6","introtext":"Sử dụng các tính chất của phép cộng, phép nhân phân số:\r\n+) Phép cộng:","display_type":2,"article_type":1},{"article_id":110659,"cat_id":941,"publish_time":1655347080,"is_link":0,"thumbnail":"","title":"Dạng 3. Thực hiện phép tính Chủ đề 6 Ôn hè Toán 6","alias":"dang-3-thuc-hien-phep-tinh-chu-de-6-on-he-toan-6","introtext":"1. Phép cộng\r\n\r\nMuốn cộng hai phân số cùng mẫu, ta cộng các tử và giữ nguyên mẫu.","display_type":2,"article_type":1},{"article_id":110658,"cat_id":941,"publish_time":1655346720,"is_link":0,"thumbnail":"","title":"Dạng 2. So sánh phân số Chủ đề 6 Ôn hè Toán 6","alias":"dang-2-so-sanh-phan-so-chu-de-6-on-he-toan-6","introtext":"Tính chất: Nếu a < b thì -a > -b\r\n\r\nCách 1: Đưa về 2 phân số có cùng mẫu số","display_type":2,"article_type":1},{"article_id":110039,"cat_id":941,"publish_time":1654766580,"is_link":0,"thumbnail":"","title":"Dạng 1. Quy đồng mẫu số các phân số Chủ đề 6 Ôn hè Toán 6","alias":"dang-1-quy-dong-mau-so-cac-phan-so-chu-de-6-on-he-toan-6","introtext":"Để quy đồng nhiều phân số, ta thường làm như sau:","display_type":2,"article_type":1}],"event_info":{"title":"Ôn tập hè Chủ đề 6. Phân số. 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