Đáp án+Giải thích các bước giải:
`\sqrt{(x-2)^2}=\sqrt{x-2}(x>=2)`
`<=>(x-2)^2=(x-2)`
`<=>(x-2)^2-(x-2)=0`
`<=>(x-2)(x-2-1)=0`
`<=>(x-2)(x-3)=0`
`<=>[(x=2),(x=3):}(tm)`
Vậy `S={2;3}`
`\sqrt{x^2-1}-\sqrt{x-1}.\sqrt{2x+1}=0(x>=1)`
`<=>\sqrt{(x-1)(x+1)}-\sqrt{x-1}.\sqrt{2x+1}=0`
`<=>\sqrt{x-1}(\sqrt{x+1}-\sqrt{2x+1})=0`
`<=>[(\sqrt{x-1}=0),(\sqrt{x+1}=\sqrt{2x+1}):}`
`<=>[(x-1=0),(x+1=2x+1):}`
`<=>[(x=1),( 2x-x+1-1=0):}`
`<=>[(x=1(tm)),(x=0(loại)):}`
Vậy `S={1}`
`\sqrt{x}+\frac{16}{\sqrt{x}}=8(x>0)`
`<=>\frac{x}{\sqrt{x}}+\frac{16}{\sqrt{x}}=8`
`<=>\frac{x+16}{\sqrt{x}}=8`
`<=>x+16=8\sqrt{x}`
`<=>x-8\sqrt{x}+16=0`
`<=>x-2.4.\sqrt{x}+4^2=0`
`<=>(\sqrt{x}-4)^2=0`
`<=>\sqrt{x}-4=0`
`<=>\sqrt{x}=4`
`<=>x=16(tm)`
Vậy `S={16}`
