a: ĐKXĐ: \(x\notin\left\{3;-3\right\}\)
\(\left(x^2-9\right):\dfrac{2x+6}{x-3}\)
\(=\left(x^2-9\right)\cdot\dfrac{x-3}{2x+6}\)
\(=\dfrac{\left(x-3\right)\left(x+3\right)\cdot\left(x-3\right)}{2\left(x+3\right)}=\dfrac{\left(x-3\right)^2}{2}\)
b: ĐKXĐ: \(\left\{{}\begin{matrix}x\notin\left\{0;\dfrac{3}{2}\right\}\\y\ne0\end{matrix}\right.\)
\(\dfrac{xy}{2x-3}:\dfrac{x^2y^2}{6-4x}\)
\(=\dfrac{xy}{2x-3}\cdot\dfrac{-4x+6}{x^2y^2}\)
\(=\dfrac{xy}{2x-3}\cdot\dfrac{-2\left(x-3\right)}{x^2y^2}\)
\(=\dfrac{-2}{xy}\)
c: ĐKXĐ: \(x\notin\left\{0;1;-2;2\right\}\)
\(\dfrac{x^2+2x}{x^2-2x+1}:\dfrac{x^2-4}{x^2-x}\)
\(=\dfrac{x\left(x+2\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2}{\left(x-1\right)\left(x-2\right)}\)
d: ĐKXĐ: \(x\ne2;y\ne-\dfrac{2x}{3}\)
\(\dfrac{2x+3y}{2-x}:\dfrac{4x^2+12xy+9y^2}{x^3-8}\)
\(=\dfrac{2x+3y}{-\left(x-2\right)}\cdot\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{4x^2+12xy+9y^2}\)
\(=\dfrac{-\left(2x+3y\right)}{1}\cdot\dfrac{x^2+2x+4}{\left(2x+3y\right)^2}\)
\(=\dfrac{-\left(x^2+2x+4\right)}{2x+3y}\)
a. Đây là 1 đề bài ko rõ ràng, ko biết là \(x^2-9:\dfrac{2x+6}{x-3}\) hay \(\left(x^2-9\right):\dfrac{2x+6}{x-3}\)
b.
\(\dfrac{xy}{2x-3}:\dfrac{x^2y^2}{6-4x}=\dfrac{xy}{2x-3}.\dfrac{-2\left(2x-3\right)}{x^2y^2}=\dfrac{-2}{xy}\)
c.
\(\dfrac{x^2+2x}{x^2-2x+1}:\dfrac{x^2-4}{x^2-x}=\dfrac{x\left(x+2\right)}{\left(x-1\right)^2}.\dfrac{x\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2}{\left(x-1\right)\left(x-2\right)}\)
d.
\(\dfrac{2x+3y}{2-x}:\dfrac{4x^2+12xy+9y^2}{x^3-8}=\dfrac{2x+3y}{-\left(x-2\right)}.\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(2x+3y\right)^2}\)
\(=-\dfrac{x^2+2x+4}{2x+3y}\)
