Question

\(A=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right):\frac{\sqrt{x}-2}{x+\sqrt{x}}\)

a) Rút gọn A

b) Tìm x để \(\sqrt{A^2}=A\)

Answer 1

a) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

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

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

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

b) Để \(\sqrt{A^2}=A\) thì \(\frac{\sqrt{x}+1}{\sqrt{x}-2}\ge0\)

Ta có : \(x>0\Rightarrow\sqrt{x}+1>0\)

\(\Rightarrow\sqrt{x}-2>0\) thì \(\sqrt{A^2}=A\)

\(\Rightarrow\sqrt{x}>2\Rightarrow x>4\)

Vậy: \(x>4\) thì \(\sqrt{A^2}=A\)

Chúc bạn học tốt!