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c) Đk: x \(\ge\)0; x \(\ne\)4; x \(\ne\)9
A = \(-\frac{1}{\sqrt{x}-3}\) => -2A = \(\frac{2}{\sqrt{x}-3}\)
Để -2A thuộc Z <=> \(2⋮\sqrt{x}-3\)
<=> \(\sqrt{x}-3\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Lập bảng:
| \(\sqrt{x}-3\) | 1 | -1 | 2 | -2 |
| x | 8 | 4 (ktm) | 25 | 1 |
Vậy ....
\(dkxd\Leftrightarrow\hept{\begin{cases}x\ge0\\\sqrt{x}-2\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}}\)
\(A=\left(\frac{\sqrt{x}}{x-4}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right):\frac{1}{\sqrt{x}+2}.\)
\(=\left(\frac{\sqrt{x}}{x-4}-\frac{2\left(\sqrt{x}+2\right)}{x-4}+\frac{\sqrt{x}-2}{x-4}\right):\frac{1}{\sqrt{x}+2}\)
\(=\frac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+2}{1}\)
\(=\frac{-6\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=-\frac{6}{\sqrt{x}-2}\)
\(A=\)\(\left(\frac{\sqrt{x}}{x-4}+\frac{2}{2-\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right)\)\(:\frac{1}{\sqrt{x}+2}\)
a,ĐKXĐ:\(\hept{\begin{cases}x\ge0\\2-\sqrt{x}\\x-4\ne0\end{cases}\ne0}\)\(\Rightarrow\)\(\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
\(A=\)\(\left(\frac{\sqrt{x}}{x-4}+\frac{2}{2-\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right)\)\(:\frac{1}{\sqrt{x}+2}\)
\(A=\)\(\left(\frac{\sqrt{x}}{x-4}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right)\)\(.\left(\sqrt{x}+2\right)\)
\(A=\)\(\left(\frac{\sqrt{x}-2\left(\sqrt{x}+2\right)+\sqrt{x}-2}{x-4}\right)\)\(.\left(\sqrt{x}+2\right)\)
\(A=\)\(\left(\frac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{x-4}\right)\)\(.\left(\sqrt{x}+2\right)\)
\(A=\)\(\left(\frac{-6}{x-4}\right)\)\(.\left(\sqrt{x}+2\right)\)
\(A=\)\(\frac{-6}{\sqrt{x}-2}\)
b,\(x=9-4\sqrt{5}\)\(\Rightarrow\)\(A=\)\(\frac{-6}{\sqrt{9-4\sqrt{5}}-2}\)\(=\frac{-6}{\sqrt{5-2.2\sqrt{5}+4}-2}\)
\(A=\)\(\frac{-6}{\sqrt{\left(\sqrt{5}-2\right)^2}-2}\)\(=\frac{-6}{\sqrt{5}-2-2}\)\(=\frac{-6}{\sqrt{5}-4}\)
c,\(A>-1\)\(\Rightarrow\)\(\frac{-6}{\sqrt{x}-2}\)\(>-1\)\(\Rightarrow\)\(\frac{-6}{\sqrt{x}-2}+1>0\)
\(\Leftrightarrow\)\(\frac{-6+\sqrt{x}-2}{\sqrt{x}-2}>0\)
\(\Leftrightarrow\)\(\frac{\sqrt{x}-8}{\sqrt{x}-2}>0\)