a) P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1+2}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\)
Để P \(\in Z\) \(\Leftrightarrow\) \(2⋮\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}-1\in U\left(2\right)\)
Bảng:
\(\sqrt{x}\) - 1 -2 2 -1 1
\(\sqrt{x}\) -1 3 0 2
x \(\varnothing\) 9 0 4
Vậy x \(\in\) {9;0;2}
a,\(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1+2}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\)
Để P nguyên thì \(\sqrt{x}-1\) thuộc \(Ư\left\{2\right\}\)
\(\Leftrightarrow\sqrt{x}-1\in\left\{1;-1;2;-2\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{-2;0;1;-3\right\}\)
mà \(\sqrt{x}\ge0\Rightarrow\sqrt{x}\in\left\{0;1\right\}\)
\(\Leftrightarrow x\in\left\{0;1\right\}\)
Vậy....
b , tương tự
b) Q = \(\dfrac{\sqrt{x}+1}{-\sqrt{x}+2}=-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}=-\dfrac{\sqrt{x}-2+3}{\sqrt{x}-2}\)
= -1 - \(\dfrac{3}{\sqrt{x}-2}\)
Để Q \(\in\) Z \(\Leftrightarrow\) 3 \(⋮\) \(\sqrt{x}-2\)
Bảng:
| \(\sqrt{x}-2\) | -1 | 1 | -3 | 3 |
| \(\sqrt{x}\) | 1 | 3 | -1 | 5 |
| x | 1 | 9 | \(\varnothing\) | 25 |
