Question

A=\(\left(\frac{3-x}{x+3}.\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)

rút gọn A

Answer 1

\(A=\left(\frac{3-x}{x+3}.\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)

\(A=\left(\frac{3-x}{x+3}.\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)

\(A=\left(\frac{-\left(x-3\right)}{x+3}.\frac{\left(x+3^2\right)}{\left(x-3\right)\left(x+3\right)}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)

\(A=\left(-1+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)

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

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

\(A=-x^2\)