Question

rút gọn

A=( \(\dfrac{\sqrt{x}}{\sqrt{x}+3}\)-\(\dfrac{x-2}{x-9}\)):\(\dfrac{3-\sqrt{x}}{x-6\sqrt{x}+9}\)

Answer 1

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
Ta có: \(A=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{(\sqrt{x}+3)\left(\sqrt{x}-3\right)}-\dfrac{x-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\dfrac{-\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)^2}\)  \(=\left[\dfrac{x-3\sqrt{x}-x+2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right].-\left(\sqrt{x}-3\right)\)

  \(=\dfrac{2-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.-\left(\sqrt{x}-3\right)\)

\(=\dfrac{3\sqrt{x}-2}{\sqrt{x}+3}\)

Vậy với \(x\ge0;x\ne9\) thì \(A=\dfrac{3\sqrt{x}-2}{\sqrt{x}+3}\)