a) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow\left(24x^2+16x-9x-6\right)-\left(4x^2+16x+7x+28\right)=10x^2-2x+5x-1\)
\(\Leftrightarrow24x^2+16x-9x-6-4x^2-16x-7x-28=10x^2-2x+5x-1\)
\(\Leftrightarrow24x^2+16x-9x-4x^2-16x-7x-10x^2+2x-5x=6+28-1\)
\(\Leftrightarrow10x^2-19x=33\)
\(\Leftrightarrow10x^2-19x+33=0\)
Phương trình vô nghiệm!!!!!!!!
b) \(4\left(x-1\right)\left(x+5\right)-\left(x+2\right)\left(x+5\right)=3\left(x-1\right)\left(x+2\right)\)
\(\Leftrightarrow4\left(x^2+5x-x-5\right)-\left(x^2+5x+2x+10\right)=3\left(x^2+2x-x-2\right)\)
\(\Leftrightarrow4x^2+20x-4x-20-x^2-5x-2x-10=3x^2+6x-3x-6\)
\(\Leftrightarrow4x^2+20x-4x-x^2-5x-2x-3x^2-6x+3x=20+10-6\)
\(\Leftrightarrow6x=24\)
\(\Leftrightarrow x=4\)
Vậy \(S=\left\{4\right\}\)