5. Simplify the expression $\left((1-x) \sqrt{\frac{1+x}{1-x}}+1-x\right) \cdot\left(\sqrt{\frac{1+x}{1-x}}-1\right)$, if $-1 \leq x<1$.

Solve each problem independently on the attached paper, and each problem on its own page.

Do not sign the sheets, write only your code.

Write your answers with a pen clearly and neatly. Each problem's solution will be graded from 0 to 6 points.

## You have $120 \mathrm{~min}$ for solving.

THE NATIONAL COMPETITION COMMITTEE WISHES YOU GREAT SUCCESS.

6th national mathematics competition for students of secondary technical and vocational schools Kranj, April 22, 2006

## PROBLEMS FOR 3RD YEAR STUDENTS
5. If we use the rule $a \cdot \sqrt{b}=\sqrt{a^{2} b}$, which holds for $a \geq 0$, in the first parenthesis, we get $((1-$ $\left.x) \sqrt{\frac{1+x}{1-x}}+1-x\right)=\left(\sqrt{\frac{(1-x)^{2}(1+x)}{1-x}}+1-x\right)$, simplify and we get $(\sqrt{(1-x)(1+x)}+1-x)$. $\left(\sqrt{\frac{1+x}{1-x}}-1\right)$. Then we remove the parentheses $\sqrt{\frac{(1-x)(1+x)(1+x)}{1-x}}+\sqrt{\frac{(1-x)(1+x)}{1-x}}-\sqrt{(1-x)(1+x)}-$ $(1-x)$, simplify again and we get $\sqrt{(1+x)^{2}}-1+x$. Considering the condition $\sqrt{a^{2}}=|a|, a \in \Re$. Since $1+x \geq 0$, the solution is: $1+x-1+x=2 x$.

We can solve the problem differently: since $\left((1-x) \sqrt{\frac{1+x}{1-x}}+1-x\right)=(1-x)\left(\sqrt{\frac{1+x}{1-x}}+1\right)$, we can write $\left((1-x) \sqrt{\frac{1+x}{1-x}}+1-x\right) \cdot\left(\sqrt{\frac{1+x}{1-x}}-1\right)=(1-x)\left(\sqrt{\frac{1+x}{1-x}}-1\right)=(1-x)\left(\left|\frac{1+x}{1-x}\right|-1\right)=$ $(1-x)\left(\frac{1+x}{1-x}-1\right)=2 x$, where we can omit the absolute value due to the assumption $-1 \leq x<1$.

1. Method: Simplification of the expression in the first parenthesis $\left((1-x) \sqrt{\frac{1+x}{1-x}}+1-x\right)=\left(\sqrt{\frac{(1-x)^{2}(1+x)}{1-x}}+1-x\right)$ and simplification.1 point Correct removal of the parenthesis

$\sqrt{\frac{(1-x)(1+x)(1+x)}{1-x}}+\sqrt{\frac{(1-x)(1+x)}{1-x}}-\sqrt{(1-x)(1+x)}-(1-x) \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .1$ point Simplification of the expression to $\sqrt{(1+x)^{2}}-1+x\left((1-x) \sqrt{\frac{1+x}{1-x}}+1-x\right) \ldots \ldots \ldots \ldots \ldots \ldots \ldots .1$ point Application of the condition $|1+x|$ 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_59669f975316a35884bfg-09.jpg?height=60&width=1639&top_left_y=2426&top_left_x=206)

Justification that there is no other solution, as $1+x$ cannot be negative ................. 1 point 2. Method:

Factoring out the common factor $(1-x)\left(\sqrt{\frac{1+x}{1-x}}+1\right)\left(\sqrt{\frac{1+x}{1-x}}-1\right) \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \ldots$ point

Simplification to $(1-x)\left(\sqrt{\left(\frac{1+x}{1-x}\right)^{2}}-1\right)$ 1 point

Considering $\sqrt{a^{2}}=|a|, a \in \Re$. . . 1 point

Writing the condition and calculation:

$(1-x)\left(\frac{1+x}{1-x}-1\right)=1+x-1+x=2 x$, since $1+x \geq 0 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$

Justification that there is no other solution, as $1+x$ cannot be negative ........................................................................

Removing the fraction under any condition

$1^{*}$ point

## Third Year