Đáp án + giải thích các bước giải:
a) `\sqrt{x^2-10x+25}=7`
`->\sqrt{(x-5)^2}=7`
`->|x-5|=7`
`->`\(\left[ \begin{array}{l}x-5=7\\x-5=-7\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=12\\x=-2\end{array} \right.\)
Vậy` S={-2;12}`
b) `\sqrt{4x-8}-1/2\sqrt{x-2}+\sqrt{9x-18}=9 (x>=2)`
`->\sqrt{4}.\sqrt{x-2}-1/2\sqrt{x-2}+\sqrt{9}.\sqrt{x-2}=9`
`->2\sqrt{x-2}-1/2\sqrt{x-2}+3\sqrt{x-2}=9`
`->9/2\sqrt{x-2}=9`
`->\sqrt{x-2}=2`
`->x-2=4`
`->x=6(TM)`
Vậy `S={6}`
c) `\sqrt{x^2-4x+4}-3=x`
`->\sqrt{(x-2)^2}=x+3 (x>=-3)`
`->|x-2|=x+3`
`->`\(\left[ \begin{array}{l}x-2=x+3\\2-x=x+3\end{array} \right.\)
`->`\(\left[ \begin{array}{l}-2=3\text{ (Vô lý)}\\-2x=1\end{array} \right.\)
`->x=-1/2 (TM)`
Vậy `S={-1/2}`
d) `x-\sqrt{4-3x}=-2 (x<=4/3)`
`->\sqrt{4-3x}=x+2 (x>=-2) `
`->4-3x=x^2+4x+4`
`->x^2+7x=0`
`->x(x+7)=0`
`->`\(\left[ \begin{array}{l}x=0\\x+7=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=0(TM)\\x=-7(KTM)\end{array} \right.\)
Vậy `S={0}`
