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

\(\dfrac{-y}{y-4}=\dfrac{--y}{4-y}=\dfrac{y}{4-y}\)

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

Câu 4:B

Câu 7: A

Câu 8: B

Câu 11: C

a: 1/x^2y=1/x^2y

3/xy=3x/x^2y

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

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

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

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

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

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

a: Ta có: \(\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)+1\)

\(=\left(x^2+9x+18\right)\left(x^2+9x+20\right)+1\)

\(=\left(x^2+9x\right)^2+38\left(x^2+9x\right)+360+1\)

\(=\left(x^2+9x\right)^2+2\cdot\left(x^2+9x\right)\cdot19+19^2\)

\(=\left(x^2+9x+19\right)^2\)

b. \(x^2+y^2+2x+2y+2\left(x+1\right)\left(y+1\right)+2\)

\(=\left(x^2+2x+1\right)+2\left(x+1\right)\left(y+1\right)+\left(y^2+2y+1\right)\)

\(=\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\)

\(=\left(x+1+y+1\right)^2=\left(x+y+2\right)^2\)

c. \(x^2-2x\left(y+2\right)+y^2+4y+4\)

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

\(=\left(x-y-2\right)^2\)

d. \(x^2+2x\left(y+1\right)+y^2+2y+1\)

\(=x^2+2x\left(y+1\right)+\left(y+1\right)^2\)

\(=\left(x+y+1\right)^2\)

Quy đồng mẫu thức các phân thức sau :

a)