Đáp án:
Giải thích các bước giải:
1) `x=6+2\sqrt{5}`
`x=(\sqrt{5}+1)^2`
`⇒ \sqrt{x}=\sqrt{5}+1`
Thay vào A ta được:
`A=\frac{\sqrt{5}+1+3}{\sqrt{5}+1-2}`
`A=\frac{9+5\sqrt{5}}{4}`
2) `B=\frac{\sqrt{x}+2}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+2}-\frac{4x}{4-x}` `ĐK: x \ge 0, x \ne 4`
`B=\frac{(\sqrt{x}+2)^2}{(\sqrt{x}-2).(\sqrt{x}+2)}-\frac{(\sqrt{x}-2)^2}{(\sqrt{x}-2).(\sqrt{x}+2)}+\frac{4x}{(\sqrt{x}-2).(\sqrt{x}+2)}`
`B=\frac{x+4\sqrt{x}+4-x+4\sqrt{x}-4+4x}{(\sqrt{x}-2).(\sqrt{x}+2)}`
`B=\frac{4x+8\sqrt{x}}{(\sqrt{x}-2).(\sqrt{x}+2)}`
`B=\frac{4\sqrt{x}(\sqrt{x}+2)}{(\sqrt{x}-2).(\sqrt{x}+2)}`
`B=\frac{4\sqrt{x}}{\sqrt{x}-2}`
3) `P=B/A`
`P=\frac{4\sqrt{x}}{\sqrt{x}-2}:\frac{\sqrt{x}+3}{\sqrt{x}-2}`
`P=\frac{4\sqrt{x}}{\sqrt{x}-2}.\frac{\sqrt{x}-2}{\sqrt{x}+3}`
`P=\frac{4\sqrt{x}}{\sqrt{x}+3}`
Ta có: `P=\frac{4\sqrt{x}}{\sqrt{x}+3}=\frac{4\sqrt{x}+12-12}{\sqrt{x}+3}=4-\frac{12}{\sqrt{x}+3}`
Để `P \in \mathbb{Z}`
`⇔ \frac{12}{\sqrt{x}+3} \in \mathbb{Z}`
`⇔ \sqrt{x}+3 \in Ư(12)`
` Ư(12)={±1;±2;±3;±4;±6;±12}`
Ta có bảng sau:
`\sqrt{x}+3 -12 -6 -4 -3 -2 -1 1 2 3 4 6 12`
`\sqrt{x} -15 -9 -7 -6 -5 -4 -2 -1 0 1 3 9`
`x ║ ║ ║ ║ ║ ║ ║ ║ 0 1 9 81`
(TM) (TM) (TM) (TM)
Vậy `x \in {0;1;9;81}` thì `P \in \mathbb{Z}`
