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![](https://rs.olm.vn/images/avt/0.png?1311)
\(Q=\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}+\dfrac{2\sqrt{x}}{x-4}\left(dk:x\ge0,x\ne4\right)\\ =\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\\ =\dfrac{2\left(2-\sqrt{x}\right)+2+\sqrt{x}-2\sqrt{x}}{4-x}\\ =\dfrac{4-2\sqrt{x}+2+\sqrt{x}-2\sqrt{x}}{4-x}\\ =\dfrac{-3\sqrt{x}+6}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\\ =\dfrac{-3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{3}{\sqrt{x}+2}\)
\(b,Q=\dfrac{6}{5}\Leftrightarrow\dfrac{3}{\sqrt{x}+2}=\dfrac{6}{5}\Rightarrow15-6\left(\sqrt{x}+2\right)=0\Rightarrow15-6\sqrt{x}-12=0\)
\(\Rightarrow-6\sqrt{x}=-3\Rightarrow\sqrt{x}=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{4}\left(tm\right)\)
Vậy \(x=\dfrac{1}{4}\)thỏa mãn đề bài.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1-\left(\frac{2}{1+2\sqrt{x}}-\frac{5\sqrt{x}}{4x-1}-\frac{1}{1-2\sqrt{x}}\right):\frac{\sqrt{x}-1}{4x+4\sqrt{x}+1}\)
\(=1-\left(\frac{2\left(1-2\sqrt{x}\right)+5\sqrt{x}-1-2\sqrt{x}}{\left(1+2\sqrt{x}\right)\left(1-2\sqrt{x}\right)}\right):\frac{\sqrt{x}-1}{\left(1+2\sqrt{x}\right)^2}\)
\(=1-\frac{1-\sqrt{x}}{\left(1+2\sqrt{x}\right)\left(1-2\sqrt{x}\right)}.\frac{\left(1+2\sqrt{x}\right)^2}{\sqrt{x}-1}=1-\frac{1+2\sqrt{x}}{1-2\sqrt{x}}=2-\frac{2}{1-2\sqrt{x}}\)
để A là số nguyên thì \(1-2\sqrt{x}\) là ước của 2 khi đó ta tìm được \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Điều kiện: x \(\ge\)0; x \(\ne\) 4;x \(\ne\) 9
\(A=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(A=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right).\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-3\right)}+\frac{\left(2\sqrt{x}+1\right).\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right).\left(\sqrt{x}-2\right)}\)
\(A=\frac{2\sqrt{x}-9-\left(x-9\right)+\left(2x-4\sqrt{x}+\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-3\right)}=\frac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-3\right)}=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-3\right)}\)
\(A=\frac{\left(\sqrt{x}+1\right).\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(A=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)
Để A nguyên thì \(\frac{4}{\sqrt{x}-3}\) nguyên <=> \(\sqrt{x}-3\) \(\in\)Ư(4) = {4;-4;2;-2;1;-1}
\(\sqrt{x}-3\) | 4 | -4 | 2 | -2 | 1 | -1 |
\(\sqrt{x}\) | 7 | -1 | 5 | 1 | 4 | 2 |
x | 49 | loại | 25 | 1 | 16 | 4 |
Đối chiếu điều kiện => x \(\in\) {49;25;1;16}
\(A=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}\left(\sqrt{x}-2\right)-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
Hay \(A=\frac{3\sqrt{x}+6-6}{\sqrt{x}+2}\)
\(=3-\frac{6}{\sqrt{x}+2}\)
Để \(A\) nguyên \(\Leftrightarrow6\) chia hết cho \(\sqrt{x}+2\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Vậy ..............................
Bổ sung xíu :))
\(Đk:\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)