a, ĐKXĐ : \(\left\{{}\begin{matrix}9-x\ne0\\x\ge0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne9\\x\ge0\end{matrix}\right.\)
Ta có : \(M=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{3-11\sqrt{x}}{9-x}\)
=> \(M=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
=> \(M=\frac{2x-6\sqrt{x}+x+\sqrt{x}+3\sqrt{x}+3+11\sqrt{x}-3}{x-9}\)
=> \(M=\frac{3x+9\sqrt{x}}{x-9}=\frac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}}{\sqrt{x}-3}\)
b, Ta có : \(x=\sqrt{\sqrt{3}-\sqrt{4-2\sqrt{3}}}\)
=> \(x=\sqrt{\sqrt{3}-\sqrt{3-2\sqrt{3}+1}}\)
=> \(x=\sqrt{\sqrt{3}-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
=> \(x=\sqrt{\sqrt{3}-\sqrt{3}+1}=\sqrt{1}=1\) ( TM )
- Thay x vào biểu thức M ta được :
\(\frac{3\sqrt{1}}{\sqrt{1}-3}=\frac{3}{-2}=-1,5\)
c, Ta có : \(M=\frac{3\sqrt{x}}{\sqrt{x}-3}=\frac{3\sqrt{x}-9+9}{\sqrt{x}-3}=3+\frac{9}{\sqrt{x}-3}\)
- Để M có giá trị là số tự nhiên
<=> \(\sqrt{x}-3\inƯ_{\left(9\right)}\)
<=> \(\sqrt{x}-3=\left\{1,-1,3,-3,9,-9\right\}\)
<=> \(\sqrt{x}=\left\{4,2,6,0,12\right\}\)
<=> \(x=\left\{0;4;16;36;144\right\}\)
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