93(1203). Construct the graph of the function:
a) $y=\left|x^{2}-4 x\right|$
b) $y=\left|x^{2}-5 x+6\right|$
c) $y=x^{2}-4|x|$
d) $y=x^{2}-5|x|+6$
Solution. a) By the definition of the absolute value of a number, we have: $y=\left|x^{2}-4 x\right|=\left\{\begin{array}{l}x^{2}-4 x, \text { if } x^{2}-4 x \geqslant 0, \text { i.e., } x \in(-\infty ; 0] \cup[4 ;+\infty), \\ -x^{2}+4 x, \text { if } x^{2}-4 x<0, \text { i.e., } x \in(0 ; 4) .\end{array}\right.$

By constructing the part of the parabola $y=x^{2}-4 x$ for $x \in(-\infty ; 0] \cup[4 ;+\infty)$ and the part of the parabola $y=-x^{2}+4 x$ for $x \in(0 ; 4)$, we obtain the graph of the function $y=\left|x^{2}-4 x\right|$ (Fig. 39).

Remark. The graph of the function $y=\left|x^{2}-4 x\right|$ can also be constructed in another way: by constructing the graph $y=x^{2}-4 x$ for all $x \in(-\infty ;+\infty)$ and reflecting the part of the graph $y=x^{2}-4 x$ that lies below the x-axis symmetrically with respect to this axis.

b) Based on the definition of the absolute value of a number, we have:

$$
y=x^{2}-4|x|=\left\{\begin{array}{l}
x^{2}-4 x, \text { if } x \geqslant 0, \\
x^{2}+4 x, \text { if } x<0 .
\end{array}\right.
$$

By constructing the part of the parabola $y=x^{2}-4 x$ for all $x \in[0 ;+\infty)$ and the part of the parabola $y=x^{2}+4 x$ for all $x \in(-\infty ; 0)$, we obtain the graph of the function $y=x^{2}-4|x|$ (Fig. 40).

![](https://cdn.mathpix.com/cropped/2024_05_21_98932225530737b82071g-209.jpg?height=640&width=536&top_left_y=271&top_left_x=383)

Fig. 39

![](https://cdn.mathpix.com/cropped/2024_05_21_98932225530737b82071g-209.jpg?height=657&width=680&top_left_y=257&top_left_x=1019)

Fig. 40

Remark. Since the function $y=x^{2}-4|x|$ is even, its graph can also be constructed in another way: by constructing the part of the graph $y=x^{2}-4 x$ for all $x \geqslant 0$ and reflecting the obtained part of the graph symmetrically with respect to the y-axis.

c) $y=\left|x^{2}-5 x+6\right|=\left\{\begin{array}{l}x^{2}-5 x+6, \text { if } x \in(-\infty ; 2] \cup[3 ;+\infty) \text {, } \\ -x^{2}+5 x-6, \text { if } x \in(2 ; 3) .\end{array}\right.$

The graph is shown in Fig. 41.

d) $y=x^{2}-5|x|+6=\left\{\begin{array}{l}x^{2}-5 x+6, \text { if } x \geqslant 0, \\ x^{2}+5 x+6, \text { if } x<0 .\end{array}\right.$

The graph is shown in Fig. 42.

$94(1204)$. Solve the system of equations:
a) $\left\{x^{2}+x y+x=14\right.$,
b) $\left\{x^{3}+x^{3} y^{3}+y^{3}=17\right.$ $\left\{y^{2}+x y+y=28\right.$ $\{x+x y+y=5$.

![](https://cdn.mathpix.com/cropped/2024_05_21_98932225530737b82071g-209.jpg?height=523&width=482&top_left_y=1823&top_left_x=387)

Fig. 41

![](https://cdn.mathpix.com/cropped/2024_05_21_98932225530737b82071g-209.jpg?height=525&width=685&top_left_y=1822&top_left_x=1005)

Fig. 42

Solution. a) Method I. We have:

$$
\left\{\begin{array}{l}
x(x+y+1)=14 \\
y(x+y+1)=28
\end{array}\right.
$$

By dividing the second equation by the first, we get the system:

![](https://cdn.mathpix.com/cropped/2024_05_21_98932225530737b82071g-210.jpg?height=157&width=888&top_left_y=498&top_left_x=578)

from which $x_{1}=-\frac{7}{3}, x_{2}=2 ; y_{1}=-\frac{14}{3}, y_{2}=4$.

Method II. After adding both equations of the system and grouping the terms, we get:

$$
(x+y)^{2}+(x+y)-42=0
$$

from which $x+y=6$ or $x+y=-7$.

In the first case, substituting $y=6-x$ into the first equation of the system, we get $x=2$, then $y=4$.

In the second case, $y=-7-x$, then $x=-2 \frac{1}{3}, y=-4 \frac{2}{3}$.

Answer: $\left(-2 \frac{1}{3} ;-4 \frac{2}{3}\right),(2 ; 4)$.
b) We have?

$$
\left\{\begin{array}{l}
\left(x^{3}+y^{3}\right)+x^{3} y^{3}=17 \\
x+y+x y=5
\end{array}\right.
$$

Since $x^{3}+y^{3}=(x+y)^{3}-3 x y(x+y)$, we have:

$$
\left\{\begin{array}{l}
(x+y)^{3}-3 x y(x+y)+x^{3} y^{3}=17 \\
(x+y)+x y=5
\end{array}\right.
$$

To simplify the solution of this system, we introduce the notation: $x+y=u, x y=v$. We have:

$$
\left\{\begin{array} { l } 
{ u ^ { 3 } - 3 u v + v ^ { 3 } = 1 7 , } \\
{ u + v = 5 ; }
\end{array} \quad \left\{\begin{array}{l}
u^{3}-3 u v+v^{3}=17 \\
u=5-v
\end{array}\right.\right.
$$

Substituting $u=5-v$ into the first equation of the system, we get the quadratic equation $v^{2}-5 v+6=0$, from which $v_{1}=2, v_{2}=3$. Therefore, $u_{1}=3, u_{2}=2$ and to determine $x$ and $y$ we have two systems:

$$
\left\{\begin{array} { l } 
{ x + y = 3 , } \\
{ x y = 2 }
\end{array} \left\{\begin{array}{l}
x+y=2 \\
x y=3
\end{array}\right.\right.
$$

The first of these systems has two solutions: $x_{1}=1, y_{1}=2$; $x_{2}=2, y_{2}=1$. The second system has no real solutions.

Answer: $(1 ; 2),(2 ; 1)$.