Question

Rút gọn biểu thức:biết -1<x<1;x≠0

Q=\(\left(\dfrac{\sqrt{1+x}}{\sqrt{1+x}-\sqrt{1-x}}+\dfrac{1-x}{\sqrt{1-x^2}-1+x}\right)\) :\(\left(\sqrt{\dfrac{1}{x}-1}-\dfrac{1}{x}\right)\)

Answer 1

\(B=\dfrac{\sqrt{1+x}}{\sqrt{1+x}-\sqrt{1-x}}+\dfrac{1-x}{\sqrt{1-x^2}-\left(1-x\right)}\)

\(=\dfrac{\sqrt{1+x}}{\sqrt{1+x}-\sqrt{1-x}}+\dfrac{1-x}{\sqrt{1-x}\left(\sqrt{1+x}-\sqrt{1-x}\right)}\)

\(=\dfrac{\sqrt{1-x^2}+1-x}{\sqrt{1-x}\left(\sqrt{1+x}-\sqrt{1-x}\right)}\)

\(=\dfrac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\)

\(\Leftrightarrow Q=\dfrac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}:\left(\dfrac{\sqrt{1-x}}{\sqrt{x}}-\dfrac{1}{x}\right)\)

\(=\dfrac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}:\dfrac{\sqrt{1-x^2}-1}{x}\)

\(=\dfrac{\left(\sqrt{1+x}+\sqrt{1-x}\right)^2}{1+x-1+x}\cdot\dfrac{x}{\sqrt{1-x^2}-1}\)

\(=\dfrac{1+x+1-x+2\sqrt{1-x^2}}{2\left(\sqrt{1-x^2}-1\right)}=1\)