Question

Rút gọn biểu thức \(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)^2\times\dfrac{x^2-1}{2}-\sqrt{1-x^2}\)

Answer 1

ĐKXĐ : \(0\le x\le1\)

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

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

\(=\dfrac{4x}{\left(x-1\right)^2}.\dfrac{\left(x-1\right)\left(x+1\right)}{2}-\sqrt{\left(1-x\right)\left(1+x\right)}\)

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

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

 

 

 

Answer 2