\(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)^2.\dfrac{x^2-1}{2-\sqrt{1-x}}\) đề rối quá :((chẳng biết thế này đúng hk
Lời giải
ĐKXĐ: \(0\le x\le1\) và \(x\ne1\)
\(A=\left(\dfrac{\sqrt{x}+1}{x-1}+\dfrac{\sqrt{x}-1}{x-1}\right)^2.\dfrac{\left(x+1\right)\left(x-1\right)}{2-\sqrt{1-x}}\)
\(A=\left(\dfrac{\sqrt{x}+1+\sqrt{x}-1}{x-1}\right)^2.\dfrac{\left(x+1\right)\left(x-1\right)}{2-\sqrt{1-x}}\)
\(A=\dfrac{4x}{\left(x-1\right)^2}.\dfrac{\left(x+1\right)\left(x-1\right)}{2-\sqrt{1-x}}\)
\(A=\dfrac{4x\left(x+1\right)}{\left(x-1\right)\left(2-\sqrt{1-x}\right)}\)