\(2\left|3x+1\right|+1=5\)
=> \(2\left|3x-1\right|=5-1=4\)
=> \(\left|3x-1\right|=4:2=2\)
=> \(\left[{}\begin{matrix}3x-1=2\\3x-1=-2\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}3x=2+1=3\\3x=-2+1=-1\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=3:3=1\\x=-1:3=-\frac{1}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{1;-\frac{1}{3}\right\}\)
2 |3x - 1| + 1 = 5
=> 2 |3x - 1| = 5 - 1
=> 2 |3x - 1| = 4
=> |3x - 1| = 4 : 2
=> |3x - 1| = 2
=> \(\left\{{}\begin{matrix}3x-1=2\\3x-1=-2\end{matrix}\right.\) => \(\left\{{}\begin{matrix}3x=3\\3x=-1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=1\\x=-\frac{1}{3}\end{matrix}\right.\)
Vậy x \(\in\) \(\left\{1;-\frac{1}{3}\right\}\).
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