a) \(\frac{9x^2}{11y^2}:\frac{6x}{11y}=\frac{9x^2}{11y^2}\cdot\frac{11y}{6x}=\frac{3xy}{2}\)

b) \(\frac{x^2-49}{x-7}+x-2=\frac{\left(x-7\right)\left(x+7\right)}{x-7}+x-2=x+7+x-2=2x+5\)

c) \(\frac{3}{x+3}+\frac{1}{x-3}-\frac{18}{9-x^2}\)

= \(\frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{1\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{18}{\left(3-x\right)\left(x+3\right)}\)

= \(\frac{3x-9}{\left(x-3\right)\left(x+3\right)}+\frac{x+3}{\left(x-3\right)\left(x+3\right)}+\frac{18}{\left(x-3\right)\left(x+3\right)}\)

= \(\frac{3x-9+x+3+18}{\left(x-3\right)\left(x+3\right)}\)

= \(\frac{4x+12}{\left(x-3\right)\left(x+3\right)}\)

= \(\frac{4\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{4}{x-3}\)(đk: \(x-3\ne0\)=> \(x\ne3\))

a)  \(\frac{30x^3}{11y^2}.\frac{121y^5}{25x}=\frac{6x^2.11y^3}{5}=\frac{66x^2y^3}{5}\)

b)  \(\frac{x+3}{x^2-4}.\frac{8-12x+6x^2-x^3}{9x+27}=\frac{x+3}{\left(x-2\right)\left(x+2\right)}.\frac{\left(2-x\right)^3}{9\left(x+3\right)}\)

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

p/s: chúc bạn học tốt

f: \(=\dfrac{5x-3-x+3}{4x^2y}=\dfrac{4x}{4x^2y}=\dfrac{1}{xy}\)

g: \(=\dfrac{3x+10-x-4}{x+3}=\dfrac{2x+6}{x+3}=2\)

h: \(=\dfrac{4-2+x}{x-1}=\dfrac{x+2}{x-1}\)

n: \(=\dfrac{3x-x+6}{x\left(x+3\right)}=\dfrac{2\left(x+3\right)}{x\left(x+3\right)}=\dfrac{2}{x}\)

p: \(=\dfrac{x^2-9-x^2+9}{x\left(x-3\right)}=0\)

k: \(=\dfrac{x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{-6}{x^2-4}\)

m: \(=\dfrac{3x-x+6}{x\left(2x+6\right)}=\dfrac{2x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)

Điều kiện: \(\hept{\begin{cases}3\left(x+y\right)\ne0\\x^2-2xy+y^2\ne0\\6\left(x+y\right)\ne0\end{cases}\Rightarrow}\hept{\begin{cases}x+y\ne0\\\left(x-y\right)^2\ne0\\x+y\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne-y\\x\ne y\end{cases}}}\)

\(\frac{2x^3-2y^3}{3x+3y}:\frac{x^2-2xy+y^2}{6x+6y}\)

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

\(=\frac{2\left(x-y\right)\left(x^2+xy+y^2\right)}{3\left(x+y\right)}.\frac{6\left(x+y\right)}{\left(x-y\right)^2}\)

\(=\frac{4\left(x^2+xy+y^2\right)}{x-y}\)

DK: \(x\ne0;y\ne0\)

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

a) \(\frac{2x-7}{10x-4}-\frac{3x+5}{4-10x}\)

\(=\frac{2x-7}{10x-4}-\frac{-\left(3x+5\right)}{-\left(4-10x\right)}\)

\(=\frac{2x-7}{10x-4}-\frac{5-3x}{10x-4}\)

\(=\frac{2x-7-\left(5-3x\right)}{10x-4}\)

\(=\frac{2x-7-5+3x}{10x-4}\)

\(=\frac{5x-12}{10x-4}\)

cái chỗ \(\frac{2x-4x}{3}\) sai đề