Question

Cho biểu thức: A= \(\dfrac{2\sqrt{9}}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)

a) Rút gọn A

b) Tìm X để A<1

mn giải chi tiết giúp mk vs ạ! mk đang cần gấp

Answer 1

a, đk x >= 0 ; x khác 4 ; 9 

\(A=\dfrac{6-x+9+\left(2\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{x-5\sqrt{x}+6}=\dfrac{-x+15+2x+7\sqrt{x}+3}{x-5\sqrt{x}+6}\)

\(=\dfrac{x-7\sqrt{x}+18}{x-5\sqrt{x}+6}\)

b, Ta có \(\dfrac{x-7\sqrt{x}+18-x+5\sqrt{x}-6}{x-5\sqrt{x}+6}=\dfrac{-2\sqrt{x}+12}{x-5\sqrt{x}+6}=\dfrac{-2\left(\sqrt{x}-6\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}< 0\)

\(\Leftrightarrow\dfrac{\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}>0\)

TH1 : \(\left\{{}\begin{matrix}x>36\\\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)>0\end{matrix}\right.\Leftrightarrow x>36\)

TH2 : \(\left\{{}\begin{matrix}x< 36\\x< 4\end{matrix}\right.\Leftrightarrow x< 4\)

Answer 2

DKXD: \(\left\{{}\begin{matrix}x\ge0\\x-5\sqrt{x}+6\ne0\\\sqrt{x}-2\ne0\\3-\sqrt{x}\ne0\end{matrix}\right.\)  \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne2\\\sqrt{x}\ne3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne4\\x\ne9\end{matrix}\right.\)

a, 

 \(A=\dfrac{2\sqrt{9}}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)

\(=\dfrac{2\sqrt{9}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\)

\(=\dfrac{2\sqrt{9}-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{2\sqrt{9}-\left(x-9\right)+2x+\sqrt{x}-4\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x-3\sqrt{x}-11+2\sqrt{9}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

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