Question

1. cho A= \(\dfrac{\sqrt{X}}{\sqrt{X}-5}\) - \(\dfrac{10\sqrt{x}}{x-25}\) - \(\dfrac{5}{\sqrt{X}+5}\) ( x>0 ; x\(\ne\)15)

a) rút gọn và tính A khi x=9

b) tìm x để A < \(\dfrac{1}{3}\)

2. Cho A = \(\dfrac{\sqrt{X}}{\sqrt{X}+3}\) + \(\dfrac{2\sqrt{X}}{\sqrt{X}-3}\) - \(\dfrac{3X+9}{X-9}\)

a) rút gọn A 

b) tìm X để A=\(\dfrac{1}{3}\)

c) tìm giá trị lớn nhất của A

Answer 1

1.

\(a.A=\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\left(x\ge0;x\ne25\right)\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+5\right)}{x-25}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5\left(\sqrt{x}-5\right)}{x-25}\)

\(=\dfrac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{x-25}\)

\(=\dfrac{x-10\sqrt{x}+25}{x-25}\)

\(=\dfrac{\left(\sqrt{x}-5\right)^2}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)

\(=\dfrac{\sqrt{x}-5}{\sqrt{x+5}}\)

\(=\dfrac{\sqrt{9}-5}{\sqrt{9}+5}=\dfrac{-2}{8}=-\dfrac{1}{4}\)

 

Answer 2

1. b.

\(\dfrac{\sqrt{x}-5}{\sqrt{x}+5}< \dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{3\left(\sqrt{x}-5\right)}{3\left(\sqrt{x}+5\right)}-\dfrac{\sqrt{x}+5}{3\left(\sqrt{x}+5\right)}< 0\)

\(\Leftrightarrow\dfrac{3\sqrt{x}-15-\sqrt{x}-5}{3\left(\sqrt{x}+5\right)}< 0\)

\(\Leftrightarrow\dfrac{2\sqrt{x}-20}{3\left(\sqrt{x}+5\right)}< 0\)

\(\Leftrightarrow\dfrac{2\left(\sqrt{x}-10\right)}{3\left(\sqrt{x}+5\right)}< 0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2\left(\sqrt{x}-10\right)>0\\3\left(\sqrt{x}+5\right)< 0\end{matrix}\right.\\\left\{{}\begin{matrix}2\left(\sqrt{x}-10\right)< 0\\3\left(\sqrt{x}+5\right)>0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\sqrt{10}\\x< -\sqrt{5}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \sqrt{10}\\x>-\sqrt{5}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< \sqrt{10}\\x>-\sqrt{5}\end{matrix}\right.\)