\(\left|3x +1\right|=5+6x\) (1)
+ Nếu \(3x+1\ge0\Leftrightarrow x\ge-\dfrac{1}{3}\)
\(\left(1\right)\Leftrightarrow3x+1=5+6x\)
\(\Leftrightarrow-6x+3x=5-1\)
\(\Leftrightarrow-3x=4\)
\(\Leftrightarrow x=-\dfrac{4}{3}\)(loại )
+Nếu \(3x+1< 0\Leftrightarrow x< \dfrac{1}{3}\)
\(\left(1\right)\Leftrightarrow3x+1=-5-6x\)
\(\Leftrightarrow6x+3x=-5-1\)
\(\Leftrightarrow9x=-6\)
\(\Leftrightarrow x=-\dfrac{2}{3}\)(nhận)
Vậy S =\(\left\{-\dfrac{2}{3}\right\}\)
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\(\left\{{}\begin{matrix}3x+1=5+6x\\3x+1=-5-6x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-4}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
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