Bài 3:
a) `\sqrt{3x+1}=4`
`⇔3x+1=16`
`⇔3x=15`
`⇔x=5`
Vậy: `S={5}`
b) `\sqrt{x^2-10x+25}=12`
`⇔\sqrt{(x-5)^2}=12`
`⇔|x-5|=12`
`⇔`\(\left[ \begin{array}{l}x-5=12\\x-5=-12\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=17\\x=-7\end{array} \right.\)
Vậy: `S={17;-7}`
c) `3\sqrt{4x+8}-\sqrt{25x+50}+\sqrt{x+2}=2` (Điều kiện: `x≥-2`)
`⇔6\sqrt{x+2}-5\sqrt{x+2}+\sqrt{x+2}=2`
`⇔2\sqrt{x+2}=2`
`⇔\sqrt{x+2}=1`
`⇔x+2=1`
$⇔x=-1\ (\text{thỏa mãn điều kiện})$
Vậy: `S={-1}`
