\(2.\left(-x+5\right)-\dfrac{3}{2}.\left(x-4\right)=0\)
<=> \(-2x+10-\dfrac{3x}{2}+6=0\)
<=> \(\dfrac{-7x}{2}+16=0\)
<=> \(\dfrac{-7x}{2}=-16\)
<=> \(x=\dfrac{32}{7}\)
vậy ...
Ta có: \(2\left(-x+5\right)-\dfrac{3}{2}\left(x-4\right)=0\)
\(\Leftrightarrow-2x-10-\dfrac{3}{2}x+6=0\)
\(\Leftrightarrow x\cdot\dfrac{-7}{2}=4\)
hay \(x=-\dfrac{8}{7}\)