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

\(\frac{2x-1}{x^2-y^2}=\frac{2x+1}{\left(x+y\right)\left(x-y\right)}\)

\(\frac{1}{x+y}\)  giữ nguyên

MTC: \(\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)

Các nhân tử phụ tương ứng là : \(\left(x+y\right);\left(x-y\right)\left(x^2+xy+y^2\right);\left(x^2+xy+y^2\right)\)

Ta có:

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

\(\frac{1}{x+y}=\frac{1.\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)}\)

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

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

\(\frac{5}{x^2-16}=\frac{5}{\left(x-4\right)\left(x+4\right)}=\frac{5x+10}{\left(x+2\right)\left(x-4\right)\left(x+4\right)}\)

\(\frac{1}{x^2-2x-8}=\frac{1}{x\left(x-4\right)+2\left(x-4\right)}=\frac{1}{\left(x-4\right)\left(x+2\right)}=\frac{x+4}{\left(x+2\right)\left(x+4\right)\left(x-4\right)}\)

\(\dfrac{2}{x^3-y^3}=\dfrac{2}{\left(x-y\right)\left(x^2+xy+y^2\right)};\dfrac{1}{x^2-y^2}=\dfrac{1}{\left(x-y\right)\left(x+y\right)}\)MTC: (x-y)((x+y)(x²+xy+y²)

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

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

\(\frac{x+2}{4x-x^2-3}=\frac{-\left(x+2\right)}{x^2-4x+3}=\frac{\left(-x-2\right)\left(2x+5\right)}{\left(x-1\right)\left(x-3\right)\left(2x+5\right)}=\frac{-2x^2-9x-10}{\left(x-1\right)\left(x-3\right)\left(2x+5\right)}\)

\(\frac{1}{2x^2+3x-5}=\frac{1}{\left(x-1\right)\left(2x+5\right)}=\frac{x-3}{\left(x-1\right)\left(x-3\right)\left(2x+5\right)}\)

a: \(\dfrac{1}{6x^2y^3}=\dfrac{7x^2}{42x^4y^3}\)

\(\dfrac{-5}{21xy^2}=\dfrac{-10x^3y}{42x^4y^3}\)

\(\dfrac{3}{14x^4y}=\dfrac{3\cdot3y}{42x^4y^3}=\dfrac{9y}{42x^4y^3}\)

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

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