(2x+1)(3x-2)=(5x-8)(2x+1)
<=> (2x+1)(3x-2)-(5x-8)(2x+1)=0
<=>(2x+1)(3x-2-5x+8)=0
<=>(2x+1)(6-2x)=0
<=>2x+1=0 hoặc 6-2x=0
<=>x=-1/2 hoặc x=3
Vậy phương trình có tập nghiệm S={-1/2;3}
(2x+1)(3x-2)=(5x-8)(2x+1)
<=> (2x+1)(3x-2)-(5x-8)(2x+1)=0
<=>(2x+1)(3x-2-5x+8)=0
<=>(2x+1)(6-2x)=0
<=>2x+1=0 hoặc 6-2x=0
<=>x=\(-\frac{1}{2}\) hoặc x=3
Vậy phương trình có tập nghiệm \(S=\left\{\frac{-1}{2};3\right\}\)
\(\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\)
\(\Leftrightarrow6x^2-x-2=10x^2-11x-8\)
\(\Leftrightarrow6x^2-x-2-10x^2+11x+8=0\)
\(\Leftrightarrow-2\left(2x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=3\end{cases}}\)
\(\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\)
\(6x^2-x-2=10x^2-11x-8\)
\(6x^2-x-2-10x^2+11x+8=0\)
\(2\left(-2x-1\right)\left(x-3\right)=0\)
\(Th1:2\left(-2x-1\right)=0\Leftrightarrow-4x-2=0\)
\(\Leftrightarrow-4x=2\Leftrightarrow x=-\frac{1}{2}\)
\(Th2:x-3=0\Leftrightarrow x=3\)