Đáp án:
Giải thích các bước giải:
Bài 5:
`A=(\frac{\sqrt{a}+2}{\sqrt{a}+1}-\frac{\sqrt{a}-2}{\sqrt{a}-1}):\frac{1}{\sqrt{a}-1}`
ĐK: ` a > 0, a \ne 1`
a) `A=[\frac{(\sqrt{a}+2)(\sqrt{a}-1)}{(\sqrt{a}-1)(\sqrt{a}+1)}-\frac{(\sqrt{a}-2)(\sqrt{a}+1)}{(\sqrt{a}-1)(\sqrt{a}+1)}].\frac{\sqrt{a}-1}{1}`
`A=[\frac{a+\sqrt{a}-2}{(\sqrt{a}-1)(\sqrt{a}+1)}-\frac{a-\sqrt{a}-2}{(\sqrt{a}-1)(\sqrt{a}+1)}].\frac{\sqrt{a}-1}{1}`
`A=[\frac{a+\sqrt{a}-2-a+\sqrt{a}+2}{(\sqrt{a}-1)(\sqrt{a}+1)}].\frac{\sqrt{a}-1}{1}`
`A=\frac{2\sqrt{a}}{(\sqrt{a}-1)(\sqrt{a}+1)}.\frac{\sqrt{a}-1}{1}`
`A=\frac{2\sqrt{a}}{\sqrt{a}+1}`
b) `A=\frac{2\sqrt{a}}{\sqrt{a}+1}=\frac{2\sqrt{a}+2-2}{\sqrt{a}+1}=\frac{2\sqrt{a}+2}{\sqrt{a}+1}-\frac{2}{\sqrt{a}+1}=2-\frac{2}{\sqrt{a}+1}=-2+\frac{2}{\sqrt{a}+1}`
Để `A \in \mathbb{Z}`
`⇔ \frac{2}{\sqrt{a}+1} \in \mathbb{Z}`
`⇔ \sqrt{a}+1 \in Ư(2)`
` Ư(2)={±1;±2}`
Ta có bảng sau:
`\sqrt{a}+1` | -2 | -1 | 1 | 2 |
`\sqrt{a}` | -3 | -1 | 0 | 1 |
`a` | ║ | ║ | 0 | 1 |
(TM) (L)
Vậy `a=0` thì `A \in \mathbb{Z}`
