\(\left(3x+1\right)^2:8=2\)
\(\left(3x+1\right)^2=16\)
\(\left(3x+1\right)^2=4^2\)
\(\Rightarrow\orbr{\begin{cases}3x+1=4\\3x+1=-4\end{cases}}\Rightarrow\orbr{\begin{cases}3x=3\\3x=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{3}\end{cases}}\)
Vậy x = 1 hoặc \(x=-\frac{5}{3}\) .
(3x+1)^2:8=2
(3x+1)^2=8/2
(3x+1)^2=4
(3x+1)^2=4^2
(3x+1)=4
(3x)=4-1
(3x)=3
x=3/3
x=1