\(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow24x^2+7x-6-\left(4x^2+23x+28\right)=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34=10x^2+3x-1\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\Leftrightarrow10x^2-30x+11x-33=0\)
\(\Leftrightarrow10x\left(x-3\right)+11\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(10x+11\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\10x+11=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-\frac{11}{10}\end{cases}}\)
Vậy \(x\in\left\{3;-\frac{11}{10}\right\}.\)
Bài làm :
Ta có :
\(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow24x^2+7x-6-\left(4x^2+23x+28\right)=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34=10x^2+3x-1\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\Leftrightarrow10x^2-30x+11x-33=0\)
\(\Leftrightarrow10x\left(x-3\right)+11\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(10x+11\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\10x+11=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-\frac{11}{10}\end{cases}}\)
Vậy x=3 hoặc x=-11/10
