a) điều kiện xác định : \(x>1\)
b) ta có : \(A=\left(\dfrac{1}{\sqrt{x-1}}+\dfrac{1}{\sqrt{x+1}}\right)^2.\dfrac{x^2-1}{2}-\sqrt{x^2-1}\)
\(\Leftrightarrow A=\left(\dfrac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x^2-1}}\right)^2.\dfrac{x^2-1}{2}-\sqrt{x^2-1}\)
\(\Leftrightarrow A=\dfrac{2x+2\sqrt{x^2-1}}{x^2-1}.\dfrac{x^2-1}{2}-\sqrt{x^2-1}\)
\(\Leftrightarrow A=\dfrac{2x+2\sqrt{x^2-1}}{2}-\sqrt{x^2-1}=\dfrac{2x}{2}=x\)
b) ta có : \(A=2\sqrt{x}\Leftrightarrow x=2\sqrt{x}\Leftrightarrow x-2\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(L\right)\\x=4\left(N\right)\end{matrix}\right.\)
vậy \(x=4\)
