Question

\(\frac{x+9y}{x^2-9y^2}-\frac{3y}{x^2+3xy}\)

Answer 1

đk:   \(x\ne0\);  \(x\ne\pm3y\)

\(\frac{x+9y}{x^2-9y^2}-\frac{3y}{x^2+3xy}\)

\(=\frac{x+9y}{\left(x-3y\right)\left(x+3y\right)}-\frac{3y}{x\left(x+3y\right)}\)

\(=\frac{x\left(x+9y\right)}{x\left(x-3y\right)\left(x+3y\right)}-\frac{3y\left(x-3y\right)}{x\left(x-3y\right)\left(x+3y\right)}\)

\(=\frac{x^2+9xy-3xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}\)

\(=\frac{\left(x+3y\right)^2}{x\left(x-3y\right)\left(x+3y\right)}\)

\(=\frac{x+3y}{x\left(x-3y\right)}\)