Question

B=\(\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right)\): \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

Rút gọn

Answer 1

ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x< >1\end{matrix}\right.\)

\(B=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}-1}\right)\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

\(=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{1+\sqrt{x}}=\dfrac{1}{\sqrt{x}}\)

Answer 2