Question

Solve:
$\frac{\sqrt{2-x}+\sqrt{2+x}}{\sqrt{2-x}-\sqrt{2+x}}=3$, then the answer is $\frac{-6}{5}$.
If true then enter $1$ and if false then enter $0$.

Answer 1

$\frac{\sqrt{2-x}+\sqrt{2+x}}{\sqrt{2-x}-\sqrt{2-x}}=3$
$=\frac{\sqrt{2-x}+\sqrt{2+x}}{\sqrt{2-x}-\sqrt{2+x}}\times \frac{\sqrt{2-x}+\sqrt{2+x}}{\sqrt{2-x}+\sqrt{2+x}}$
$=\frac{2-x+2+x+2\sqrt{2-x}\sqrt{2+x}}{2-x-2-x}$
$=\frac{2+\sqrt{2-x}\sqrt{2+x}}{-x}=3$
$=\sqrt{2-x}\sqrt{2+x}$
$=-3x-2$.
On squaring,$=(2-x)(2+x)=(3x+2)(3x+2)$
$4-x^2=9x^2+12x+4$
$10x^2+12x=0$
$\therefore x=-\frac{6}{5}$.

The statement is true.

Therefore, the answer is $1$.