Ta có: \(\frac{b^2-3b}{2b^2-3b-9}=\frac{b^2+3b}{A}\)
\(\Leftrightarrow A=\frac{\left(b^2+3b\right)\cdot\left(2b^2-3b-9\right)}{b^2-3b}\)
\(\Leftrightarrow A=\frac{b\left(b+3\right)\left(2b^2-6b+3b-9\right)}{b\left(b-3\right)}\)
\(\Leftrightarrow A=\frac{\left(b+3\right)\left[2b\left(b-3\right)+3\left(b-3\right)\right]}{b-3}\)
\(\Leftrightarrow A=\frac{\left(b+3\right)\left(b-3\right)\left(2b+3\right)}{b-3}\)
\(\Leftrightarrow A=\left(b+3\right)\left(2b+3\right)\)
\(\Leftrightarrow A=2b^2+3b+6b+9\)
\(\Leftrightarrow A=2b^2+9b+9\)
