\(\left(3x+1\right)^2=4\\ \Leftrightarrow\left(3x+1\right)^2=\left(\pm2\right)^2\\ \Rightarrow3x+1\in\left\{2;-2\right\}\\ \Rightarrow\left[{}\begin{matrix}3x+1=2\\3x+1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=-1\end{matrix}\right.\)
Vậy...
\(\left(3x+1\right)^2=4\Leftrightarrow3x+1=2\)
\(\Leftrightarrow3x=1\Leftrightarrow x=\frac{1}{3}\)
ta có:
(3x+1)^2=4
<=>(3x+1)^2=(2 hoặc-2)
<=>3x+1 thuộc {2;-2}
<=>3x+1=2
3x+1=-2
<=>x=1/3
x=-1
vậy x= 1/3;x=-1
a) \(\left(3x+1\right)^2=4\)
\(\Rightarrow3x+1=\pm2.\)
\(\Rightarrow\left[{}\begin{matrix}3x+1=2\\3x+1=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1:3\\x=\left(-3\right):3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=-1\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{3};-1\right\}.\)
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