Đáp án+Giải thích các bước giải:
1,
a,
`\sqrt{3x^2}=\sqrt{12}`
`⇔3x^2=12`
`⇔x^2=4`
`⇔x=±2`
Vậy `S={2;-2}`
b,
`\sqrt{(x-2)^2}=3`
`⇔|x-2|=3`
$⇔\left[\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.$
$⇔\left[\begin{matrix}x=5\\x=-1\end{matrix}\right.$
Vậy `S={5;-1}`
c,
`\sqrt{4(x^2+6x+9)}=8`
`⇔2.\sqrt{(x+3)^2}=8`
`⇔|x+3|=4`
$⇔\left[\begin{matrix}x+3=4\\x+3=-4\end{matrix}\right.$
$⇔\left[\begin{matrix}x=1\\x=-7\end{matrix}\right.$
Vậy `S={1;-7}`
d,
`\sqrt{3x^2-6x+3}=\sqrt{3}`
`⇔3x^2-6x+3=3`
`⇔3x^2-6x=0`
`⇔3x(x-2)=0`
$⇔\left[\begin{matrix}3x=0\\x-2=0\end{matrix}\right.$
$⇔\left[\begin{matrix}x=0\\x=2\end{matrix}\right.$
Vậy `S={0;2}`
