\(\dfrac{3x}{\left|x-1\right|}=\dfrac{x-2}{x}\Leftrightarrow3x.x=\left(x-2\right).\left|x-1\right|\)
\(\Leftrightarrow3x^2=\left(x-2\right).\left|x-1\right|\)
th1: \(x-1\ge0\Leftrightarrow x\ge1\)
\(\Rightarrow3x^2=\left(x-2\right).\left|x-1\right|\Leftrightarrow3x^2=\left(x-2\right)\left(x-1\right)\)
\(\Leftrightarrow3x^2=x^2-x-2x+2\Leftrightarrow3x^2-x^2+x+2x-2=0\)
\(\Leftrightarrow2x^2-x+4x-2=0\Leftrightarrow x\left(2x-1\right)+2\left(2x-1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x+2=0\\2x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-2\\2x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(loại\right)\\x=\dfrac{1}{2}\left(loại\right)\end{matrix}\right.\)
th2: \(x-1< 0\Leftrightarrow x< 1\)
\(\Rightarrow3x^2=\left(x-2\right).\left|x-1\right|\Leftrightarrow3x^2=\left(x-2\right)\left(1-x\right)\)
\(\Leftrightarrow3x^2=x-x^2-2+2x\Leftrightarrow3x^2-x+x^2+2-2x=0\)
\(\Leftrightarrow4x^2-3x+2=0\Leftrightarrow\left(4x^2-3x+\dfrac{9}{16}\right)+\dfrac{23}{16}=0\)
\(\Leftrightarrow\left(2x-\dfrac{3}{4}\right)^2+\dfrac{23}{16}=0\)
ta có : \(\left(2x-\dfrac{3}{4}\right)^2\ge0\) với mọi \(x\) \(\Rightarrow\left(2x-\dfrac{3}{4}\right)^2+\dfrac{23}{16}\ge\dfrac{23}{16}>0\) với mọi \(x\)
vậy phương trình vô nghiệm
