a) \(3x\left(2x+3\right)-\left(2x+5x\right)\left(3x-2\right)=8\)

\(\Leftrightarrow6x^2+9x-\left(6x^2-4x+15x^2-10x\right)=8\)

\(\Leftrightarrow6x^2+9x-6x^2+4x-15x^2+10x=8\)

\(\Leftrightarrow23x-15x^2-8=0\)

\(\Leftrightarrow-15x\left(x-1\right)+8\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(8-15x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\8-15x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{8}{15}\end{cases}}}\)

Vậy ....

\(3x\left(2x+3\right)-\left(2x+5x\right)\left(3x-2\right)=8\)

\(\Leftrightarrow6x^2+9x-\left(6x^2-4x+15x^2-10x\right)=8\)

\(\Leftrightarrow6x^2+9x-6x^2+4x-15x^2+10x=8\)

\(\Leftrightarrow-15x^2+23x-8=0\)

\(\Leftrightarrow-\left(15x^2-23x+8\right)=0\)

\(\Leftrightarrow-\left(15x^2-15x-8x+8\right)=0\)

\(\Leftrightarrow-\left(15x-8\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(1-x\right)\left(15x-8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}1-x=0\\15x-8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{8}{15}\end{cases}}}\)

a, \(3x\left(2x+3\right)-\left(2x+5x\right)\left(3x-2\right)=8\)

\(\Leftrightarrow6x^2+9x-21x^2+14x=8\)

\(\Leftrightarrow-15x^2+23x-8=0\)

\(\Leftrightarrow x\left(15x-8\right)-\left(15x-8\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(15x-8\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{8}{15}\end{cases}}\)