\(\left(x-2\right)\left(2x+1\right)-5\left(x+3\right)=2x\left(x-3\right)+4\left(1+2x\right)-2\left(1+x\right)\)
\(2x^2+x-4x-2-5x-15=2x^2-6x+4+8x-2-2x\)
\(x-4x-2-5x-15=-6x+4+8x-2-2x\)
\(\Rightarrow-8x-17=2\)
\(-8x=19\Rightarrow x=-\dfrac{19}{8}\)
Vậy \(x=-\dfrac{19}{8}\)
\(\left(x-2\right)\left(2x+1\right)-5\left(x+3\right)=2x\left(x-3\right)+4\left(1+2x\right)-2\left(1+x\right)\)
\(\Rightarrow2x^2+x-4x-2-5x-15-2x^2+6x-4-8x+2+2x=0\\ \Leftrightarrow-8x-19=0\\ \Leftrightarrow-8x=19\\ \Leftrightarrow x=-\dfrac{19}{8}\)