`2 (2x-1) -x=4`
`<=> 4x - 2-x=4`
`<=> 3x -2 =4`
`<=> 3x-2-4=0`
`<=> 3x-6=0`
`<=> 3x=6`
`<=>x=2`
Vậy pt có nghiệm `x=2`
__
\(\dfrac{x+1}{x-2}+\dfrac{1}{x}=\dfrac{5x-4}{x^2-2x}\)
\(\Leftrightarrow\dfrac{x+1}{x-2}+\dfrac{1}{x}=\dfrac{5x-4}{x\left(x-2\right)}\)
\(ĐKXĐ:\left\{{}\begin{matrix}x\ne0\\x-2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne2\end{matrix}\right.\)
Ta có : \(\dfrac{x+1}{x-2}+\dfrac{1}{x}=\dfrac{5x-4}{x\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x\left(x+1\right)}{x\left(x-2\right)}+\dfrac{x-2}{x\left(x-2\right)}=\dfrac{5x-4}{x\left(x-2\right)}\)
`=> x^2 + x + x-2 -5x+4=0`
`<=> x^2 -3x +2=0`
`<=> x^2 -x-2x+2=0`
`<=> x(x-1) -2(x-1)=0`
`<=> (x-1)(x-2)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\left(l\right)\end{matrix}\right.\)
Vậy pt có nghiệm `x=1``
2(2x-1)-x=4
\(\Leftrightarrow\) 4x-2-x=4
\(\Leftrightarrow\) 4x-x=4+2
\(\Leftrightarrow\) 3x=6
\(\Leftrightarrow\) x= 2
