a:
sửa đề: \(Q=\dfrac{x^2+3x}{x^2-9}\)
ĐKXĐ: \(x\notin\left\{0;-3;3\right\}\)
\(P=\dfrac{x^2+6x+9}{x^2+3x}=\dfrac{\left(x+3\right)^2}{x\left(x+3\right)}=\dfrac{x+3}{x}\)
\(Q=\dfrac{x^2+3x}{x^2-9}=\dfrac{\left(x+3\right)\cdot x}{\left(x+3\right)\left(x-3\right)}=\dfrac{x}{x-3}\)
b: \(P\cdot Q=\dfrac{x+3}{x}\cdot\dfrac{x}{x-3}=\dfrac{x+3}{x-3}\)
\(P:Q=\dfrac{x+3}{x}:\dfrac{x}{x-3}=\dfrac{x+3}{x}\cdot\dfrac{x-3}{x}=\dfrac{x^2-9}{x^2}\)