a) \(Q=\) \(\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\left(x>0;x\ne1\right)\)
\(Q=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(Q=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(Q=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(Q=\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(Q=\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\) \(=\dfrac{2}{x-1}\) \(\left(đpcm\right)\).
b) Để \(Q\in Z\) <=> \(\dfrac{2}{x-1}\in Z\) <=> \(x-1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
Ta có bảng sau:
| x -1 | 1 | -1 | 2 | -2 |
| x | 2(TM) | 0(ko TM) | 3(TM) | -1(koTM) |
Vậy để biểu thức Q nhận giá trị nguyên thì \(x\in\left\{2;3\right\}\)
