\(\left|\dfrac{3}{2}x+\dfrac{1}{2}\right|=\left|4x-1\right|\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{2}=4x-1\\\dfrac{3}{2}x+\dfrac{1}{2}=-4x+1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x-4x=-1-\dfrac{1}{2}\\\dfrac{3}{2}x+4x=1-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}-2,5x=-1,5\\5,5x=0,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{5}\\x=\dfrac{1}{11}\end{matrix}\right.\)
Chúc bạn học tốt!!!
\(\left|\dfrac{3}{2}x+\dfrac{1}{2}\right|=\left|4x-1\right|\)
=> \(\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{2}=4x-1\\\dfrac{3}{2}x+\dfrac{1}{2}=-4x-1\end{matrix}\right.=>\left\{{}\begin{matrix}\dfrac{3}{2}x-4x=-1-\dfrac{1}{2}\\\dfrac{3}{2}x+4x=1-\dfrac{1}{2}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}-2,5x=-1,5\\5,5x=0,5\end{matrix}\right.=>\left\{{}\begin{matrix}x=\dfrac{3}{5}\\x=\dfrac{1}{11}\end{matrix}\right.\)
Vậy x \(\in\left\{\dfrac{3}{5};\dfrac{1}{11}\right\}\)
