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\(ĐKXĐ:\)
\(\hept{\begin{cases}x-9\ne0\\\sqrt{x}-2\ne0\\\sqrt{x}+3\ne0;x\ge0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne9\\x\ne4\\x\ge0\end{cases}}\)
Vậy...................................................
\(A=\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\right)\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\right):\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}-\sqrt{x}-3}{\left(\sqrt{x}+3\right)}:\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{-3}{\sqrt{x}+3}:\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{x-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{-3}{\sqrt{x}+3}:\frac{9-x+x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-3}{\sqrt{x}+3}:\frac{-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-3}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{4-x}\)
\(=\frac{3\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(=\frac{3}{\left(2+\sqrt{x}\right)}\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\\x\ne9\end{matrix}\right.\)
\(A=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\right):\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\\ =\left(\frac{\sqrt{x}-\sqrt{x}-3}{\sqrt{x}+3}\right):\left(\frac{9-x+x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\\ =\frac{-3}{\sqrt{x}+3}:\frac{4-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\frac{-3}{\sqrt{x}+3}\cdot\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\frac{3}{\sqrt{x}+2}\)
b) Ta có:
\(P=\frac{3}{\sqrt{x}+2}< 1\\ \Leftrightarrow\frac{3}{\sqrt{x}+2}-1< 0\\ \Leftrightarrow\frac{3-\left(\sqrt{x}+2\right)}{\sqrt{x}+2}< 0\\ \Leftrightarrow\frac{1-\sqrt{x}}{\sqrt{x}+2}< 0\\ \Leftrightarrow1-\sqrt{x}< 0\\ \Leftrightarrow\sqrt{x}>1\\ \Leftrightarrow x>1\)
Vậy với \(x>1;x\ne4;x\ne9\)thì P < 1
c) Để \(A\in Z\Leftrightarrow3⋮\sqrt{x}+2\Leftrightarrow\sqrt{x}+2\inƯ\left(3\right)\)
Ta có bảng sau
| \(\sqrt{x}+2\) | 1 | -1 | 3 | -3 |
| \(\sqrt{x}\) | -1 | -3 | 1 | -5 |
| \(x\) | loại | loại | 1(tm) | loại |
Vậy...................
a) \(ĐKXĐ:x>0;x\ne4\)
Ta có : \(P=\left(\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{4x}{2\sqrt{x}-x}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}\right)\)
\(=\left[\frac{\sqrt{x}.\sqrt{x}-4x}{\sqrt{x}.\left(\sqrt{x}-2\right)}\right]\cdot\frac{\sqrt{x}-2}{\sqrt{x}+3}\)
\(=\frac{-3x}{\sqrt{x}.\left(\sqrt{x}+3\right)}\)
b) Ta có : \(x-1=10-4\sqrt{6}=\left(\sqrt{6}-2\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{6}-2\right)^2+1}\)
......
Câu 1) a) ĐKXĐ \(x\ge0,\)\(x\ne4\)A=\(\frac{x+2\sqrt{x}-4}{2\left(x-4\right)}\)b) Mình chưa làm được Câu 2) a) ĐKXĐ \(x>0,\)\(x\ne4\)A=\(\frac{\sqrt{x}-1}{\sqrt{x}}\)b) Để a<\(\frac{1}{2}\)\(\Rightarrow\)\(\frac{\sqrt{x}-1}{\sqrt{x}}< \frac{1}{2}\)\(\Rightarrow x< 1\)\(\Rightarrow0< x< 1\)thỏa mãn bài toán c) Ta có A=\(\frac{\sqrt{x}-1}{\sqrt{x}}=1-\frac{1}{\sqrt{x}}\), để A \(\in Z\)\(\Rightarrow\sqrt{x}\inƯ\left(1\right)\), \(\Rightarrow x=1\)( thỏa mãn ĐK)
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