a) \(\dfrac{x}{x-y}+\dfrac{2y^2}{x^2-y^2}-\dfrac{x}{x+y}=\dfrac{x\left(x+y\right)+2y^2-x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{x^2+xy+2y^2-x^2+xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{2y^2+2xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{2y\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{2y}{x-y}\)

b) \(B=\dfrac{x}{x-2}-\dfrac{4x}{x^2-4}-\dfrac{2}{x+2}=\dfrac{x\left(x+2\right)-4x-2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+2x-4x-2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{x+2}\)

c) \(\dfrac{5}{x+1}-\dfrac{10}{-x^2+x-1}-\dfrac{15}{x^3+1}=\dfrac{5}{x+1}+\dfrac{10}{x^2-x+1}-\dfrac{15}{x^3+1}=\dfrac{5\left(x^2-x+1\right)+10\left(x+1\right)-15}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5x^2-5x+5+10x+10-15}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5x^2+5x}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5x\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5x}{x^2-x+1}\)

a: \(=x^2+2x-8-x^2-2x-1=-9\)

b: \(=\dfrac{x^2+6x+9+3x-9+2x^2-18x}{x\left(x-3\right)\left(x+3\right)}\)

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

\(A=\dfrac{x+3}{\left(x-3\right)^2}:\dfrac{12-x^2+x+x^2-9}{x\left(x-3\right)}\)

\(=\dfrac{x+3}{\left(x-3\right)^2}\cdot\dfrac{x\left(x-3\right)}{x+3}=\dfrac{x}{x-3}\)

a: \(A=\dfrac{x^2+2xy+y^2-x^2+xy+2y^2}{\left(x-y\right)\left(x+y\right)}\)

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

a: \(=\dfrac{x-z}{2}\)

b: \(=\dfrac{3x}{4y^3}\)

a) \(\dfrac{2\left(x+1\right)^2}{4x\left(x+1\right)}\left(x\ne0;x\ne-1\right)\)

\(=\dfrac{2\left(x+1\right)^2:2\left(x+1\right)}{4x\left(x+1\right):2\left(x+1\right)}\)

\(=\dfrac{x+1}{2x}\)

b) \(\dfrac{\left(8-x\right)\left(-x-2\right)}{\left(x+2\right)^2}\left(x\ne-2\right)\)

\(=\dfrac{-\left(8-x\right)\left(x+2\right)}{\left(x+2\right)^2}\)

\(=\dfrac{-\left(8-x\right)}{x+2}\)

\(=\dfrac{x-8}{x+2}\)

c) \(\dfrac{2\left(x-y\right)}{y-x}\left(x\ne y\right)\)

\(=\dfrac{2\left(x-y\right)}{-\left(x-y\right)}\)

\(=-2\)

d) \(\dfrac{\left(x+2\right)^2}{2x+4}\left(x\ne-2\right)\)

\(=\dfrac{\left(x+2\right)^2}{2\left(x+2\right)}\)

\(=\dfrac{x+2}{2}\)

ĐKXĐ: \(x\neq0;x\neq-1\)

\(\dfrac{2(x+1)^2}{4x(x+1)}=\dfrac{2(x+1)}{4x}=\dfrac{x+1}{2x}\)

$---$

ĐKXĐ: \(x\neq-2\)

\(\dfrac{(8-x)(-x-2)}{(x+2)^2}=\dfrac{-(8-x)(x+2)}{(x+2)^2}=\dfrac{x-8}{x+2}\)

$---$

ĐKXĐ: \(x\neq y\)

\(\dfrac{2(x-y)}{y-x}=\dfrac{-2(y-x)}{y-x}=-2\)

$---$

ĐKXĐ: \(x\neq-2\)

\(\dfrac{(x+2)^2}{2x+4}=\dfrac{(x+2)^2}{2(x+2)}=\dfrac{x+2}{2}\)

6:

a: ĐKXĐ: x<>0

\(\dfrac{x^3+3x^2+3x+1}{x^2+x}\)

\(=\dfrac{\left(x+1\right)^3}{x\left(x+1\right)}=\dfrac{\left(x+1\right)^2}{x}\)

b: ĐKXĐ: x<>1

\(\dfrac{x^3-3x^2+3x-1}{2x-2}\)

\(=\dfrac{\left(x-1\right)^3}{2\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{2}\)

c: ĐKXĐ: x<>-2

\(\dfrac{x^2+4x+4}{2x+4}\)

\(=\dfrac{\left(x+2\right)^2}{2\left(x+2\right)}\)

\(=\dfrac{x+2}{2}\)

d: ĐKXĐ: x<>-2

\(\dfrac{\left(x-1\right)\left(-x-2\right)}{x+2}\)

\(=\dfrac{\left(-x+1\right)\left(x+2\right)}{x+2}=-x+1\)

e: ĐKXĐ: x<>-y

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

g: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

\(\dfrac{-3x^2-6x}{4-x^2}=\dfrac{3x^2+6x}{x^2-4}\)

\(=\dfrac{3x\left(x+2\right)}{\left(x+2\right)\cdot\left(x-2\right)}=\dfrac{3x}{x-2}\)

7:

a: \(\dfrac{2}{5x^3y^2}=\dfrac{2\cdot4}{20x^3y^2}=\dfrac{8}{20x^3y^2}\)

\(\dfrac{3}{4xy}=\dfrac{3\cdot5\cdot x^2y}{20x^3y^2}=\dfrac{15x^2y}{20x^3y^2}\)

b: \(\dfrac{x}{x^2-2xy+y^2}=\dfrac{x}{\left(x-y\right)^2}\)

\(\dfrac{x}{x^2-xy}=\dfrac{x}{x\left(x-y\right)}=\dfrac{1}{x-y}=\dfrac{\left(x-y\right)}{\left(x-y\right)^2}\)

c: \(\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)

\(\dfrac{2}{2x+4}=\dfrac{2}{2\left(x+2\right)}=\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)

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

d:

\(\dfrac{2}{2x-6}=\dfrac{2}{2\left(x-3\right)}=\dfrac{1}{x-3};\dfrac{3}{3x-9}=\dfrac{3}{3\left(x-3\right)}=\dfrac{1}{x-3}\)

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

\(\dfrac{3}{3x-9}=\dfrac{1}{x-3}=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\)

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

a) Ta có: \(B=\left(\dfrac{x}{3x-9}+\dfrac{2x-3}{3x-x^2}\right)\cdot\dfrac{3x^2-9x}{x^2+6x+9}\)

\(=\left(\dfrac{x}{3\left(x-3\right)}-\dfrac{2x-3}{x\left(x-3\right)}\right)\cdot\dfrac{3x\left(x-3\right)}{\left(x+3\right)^2}\)

\(=\left(\dfrac{x^2}{3x\left(x-3\right)}-\dfrac{3\left(2x-3\right)}{3x\left(x-3\right)}\right)\cdot\dfrac{3x\left(x-3\right)}{\left(x+3\right)^2}\)

\(=\dfrac{x^2-6x+9}{3x\left(x-3\right)}\cdot\dfrac{3x\left(x-3\right)}{\left(x+3\right)^2}\)

\(=\dfrac{x^2-6x+9}{x^2+6x+9}\)

b) Ta có: \(A=\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right):\dfrac{1}{x+2}\)

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

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

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

\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{1}\)

\(=\dfrac{-6}{x-2}\)

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