Đáp án+Giải thích các bước giải:
a,
`\sqrt{(\sqrt{3}-\sqrt{5})^2}`
`=|\sqrt{3}-\sqrt{5}|`
`=\sqrt{5}-\sqrt{3}`
b,
`\sqrt{(\sqrt{7}-1)^2}+5`
`=|\sqrt{7}-1|+5`
`=\sqrt{7}-1+5`
`=\sqrt{7}+4`
c,
`2\sqrt{3}-\sqrt{(\sqrt{3}+1)^2}-\sqrt{3}`
`=\sqrt{3}-|\sqrt{3}+1|`
`=\sqrt{3}-(\sqrt{3}+1)`
`=\sqrt{3}-\sqrt{3}-1`
`=-1`
2,
a,
`\sqrt{x-9}=4(x≥9)`
`⇔x-9=16`
`⇔x=25(tm)`
Vậy `S={25}`
b,
`5\sqrt{2x+6}=5(x≥-3)`
`⇔\sqrt{2x+6}=1`
`⇔2x+6=1`
`⇔2x=-5`
`⇔x=\frac{-5}{2}(tm)`
Vậy `S={\frac{-5}{2}}`
c,
`\sqrt{x^2-2x}=1(x≤0;x≥2)`
`⇔x^2-2x=1`
`⇔x^2-2x-1=0`
`⇔x^2-2x+1-2=0`
`⇔(x-1)^2=2`
$⇔\left[\begin{matrix}x-1=\sqrt{2}\\x-1=-\sqrt{2}\end{matrix}\right.$
$⇔\left[\begin{matrix}x=\sqrt{2}+1(tm)\\x=-\sqrt{2}+1(tm)\end{matrix}\right.$
Vậy `S={1+\sqrt{2};1-\sqrt{2}}`
d,
`\sqrt{x^2-4x}=x(x=0;x≥4)`
`⇔x^2-4x=x^2`
`⇔-4x=0`
`⇔x=0(tm)`
Vậy `S={0}`
