bạn sóat lại đề bài nhé rồi mình trả lời cho !
a) \(ĐKXĐ:\)\(\hept{\begin{cases}\sqrt{x}-1\ne0\\\sqrt{x}+1\ne0\\x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ge0\end{cases}}\)
b)\(A=\left(\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+1}\right)^2.\frac{x^2-1}{2}-\sqrt{x}-x^2\)
\(=\left(\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\frac{\left(x-1\right)\left(x+1\right)}{2}-\sqrt{x}-x^2\)
\(=\frac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x+1\right)}{2}-\sqrt{x}-x^2\)
\(=\frac{2\sqrt{x}\left(x+1\right)}{2}-\sqrt{x}-x^2\)
\(=\sqrt{x}\left(x+1\right)-\sqrt{x}-x^2\)
\(=x\sqrt{x}+\sqrt{x}-\sqrt{x}-x^2\)
\(=x\sqrt{x}-x^2\)
đề vẫn có vấn đề nhé!