a, 3x+27=9
⇔ 3x=9-27
⇔3x=-18
⇔x=-18:3
⇔x=-6
b, 2x+12=3(x-7)
⇔2x+12=3x-21
⇔12+21=3x-2x
⇔33=x
⇔x=33
c, 2x²-1=49
⇔2x²=49+1
⇔2x²=50
⇔x²=50:2
⇔x²=25
⇔ \(\left[ \begin{array}{l}x²=5²\\x²=(-5)²\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=5\\x=-5\end{array} \right.\)
Vậy x ∈ { -5;5}