Bài 3:
a: Ta có: \(4\left(x+3\right)-5\left(x-1\right)=-7\)
\(\Leftrightarrow4x+12-5x+5=-7\)
\(\Leftrightarrow-x=-24\)
hay x=24
b: Ta có: \(x\cdot\left(x-5\right)-x\left(x+2\right)=9\)
\(\Leftrightarrow x^2-5x-x^2-2x=9\)
\(\Leftrightarrow x=-\dfrac{9}{7}\)