Đáp án:
Giải thích các bước giải:
1)
`H=\sqrt{5-2\sqrt{6}}-\sqrt{(\sqrt{2}+\sqrt{3})^2}`
`H=\sqrt{\frac{2(5-2\sqrt{6})}{2}}-|\sqrt{2}+\sqrt{3}|`
`H=\sqrt{\frac{10-4\sqrt{6}}{2}}-(\sqrt{2}+\sqrt{3})`
`H=\sqrt{\frac{6+4-4\sqrt{6}}{2}}-\sqrt{2}-\sqrt{3}`
`H=\sqrt{\frac{(\sqrt{6})^2+(2)^2-2.2\sqrt{6}}{2}}-\sqrt{2}-\sqrt{3}`
`H=\sqrt{\frac{(\sqrt{6}-2)^2}{2}}-\sqrt{2}-\sqrt{3}`
`H=\frac{\sqrt{6}-2}{\sqrt{2}}-\sqrt{2}-\sqrt{3}`
`H=\sqrt{3}-\sqrt{2}-\sqrt{2}-\sqrt{3}`
`H=-2\sqrt{2}`
2)
a) `29-3\sqrt{x}=2`
ĐK: `x \ge 0`
`⇔ 3\sqrt{x}=27`
`⇔ \sqrt{x}=9`
`⇔ x=81\ (TM)`
Vậy `S={81}`
b) `\sqrt{x^2-4x+4}-5=x`
`⇔ \sqrt{x^2-4x+4}=x+5`
ĐK: `x \ge -5`
`⇔ x^2-4x+4=x^2+10x+25`
`⇔ 14x=-21`
`⇔ x=-3/2\ (TM)`
Vậy `S={-3/2}`
