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\(\dfrac{1}{6x^2y^3}=\dfrac{7x^2}{42x^4y^3};\dfrac{-5}{21xy^2}=\dfrac{-10x^3y}{42x^4y^3};\dfrac{3}{14x^4y}=\dfrac{9y^2}{42x^4y^3}\)
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: \(\dfrac{2}{x^3-y^3}=\dfrac{2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{2\left(x+y\right)}{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)}\)
\(\dfrac{2x+1}{x^2-y^2}=\dfrac{\left(2x+1\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{\left(2x+1\right)\left(x^2+xy+y^2\right)}{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)}\)
\(\frac{-3}{x^2+6x+8}=\frac{-3}{x\left(x+2\right)+4\left(x+2\right)}=\frac{-3}{\left(x+2\right)\left(x+4\right)}=\frac{-3x+12}{\left(x+2\right)\left(x+4\right)\left(x-4\right)}\)
\(\frac{5}{x^2-16}=\frac{5}{\left(x-4\right)\left(x+4\right)}=\frac{5x+10}{\left(x+2\right)\left(x-4\right)\left(x+4\right)}\)
\(\frac{1}{x^2-2x-8}=\frac{1}{x\left(x-4\right)+2\left(x-4\right)}=\frac{1}{\left(x-4\right)\left(x+2\right)}=\frac{x+4}{\left(x+2\right)\left(x+4\right)\left(x-4\right)}\)
\(\dfrac{-5}{6x^5y^3}=\dfrac{-5\cdot2\cdot y^3}{12x^5y^6}=\dfrac{-10y^3}{12x^5y^6}\)
\(\dfrac{3}{4x^2y^6}=\dfrac{3\cdot3x^3}{12x^5y^6}=\dfrac{9x^3}{12x^5y^6}\)
\(\dfrac{7}{3x^4y^5}=\dfrac{7\cdot4\cdot x\cdot y}{12x^5y^6}=\dfrac{28xy}{12x^5y^6}\)
\(\frac{x+2}{4x-x^2-3}=\frac{-\left(x+2\right)}{x^2-4x+3}=\frac{\left(-x-2\right)\left(2x+5\right)}{\left(x-1\right)\left(x-3\right)\left(2x+5\right)}=\frac{-2x^2-9x-10}{\left(x-1\right)\left(x-3\right)\left(2x+5\right)}\)
\(\frac{1}{2x^2+3x-5}=\frac{1}{\left(x-1\right)\left(2x+5\right)}=\frac{x-3}{\left(x-1\right)\left(x-3\right)\left(2x+5\right)}\)
MTC=(x+2)(x+4)(x-2)(x-4)
\(\dfrac{-3}{x^2+6x+8}=\dfrac{-3\left(x-2\right)\left(x-4\right)}{\left(x+2\right)\left(x-2\right)\left(x+4\right)\left(x-4\right)}\)
\(\dfrac{5}{x^2-16}=\dfrac{5\left(x+2\right)\left(x-2\right)}{\left(x-4\right)\left(x+4\right)\left(x-2\right)\left(x+2\right)}\)
\(\dfrac{1}{x^2-2x-8}=\dfrac{1}{\left(x-4\right)\left(x+2\right)}=\dfrac{\left(x+4\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)\left(x+4\right)\left(x-4\right)}\)
\(\frac{1}{6x^2y^3}=\frac{7x^2}{42x^4y^3},\frac{-5}{21xy^2}=\frac{-10x^3y}{42x^4y^3},\frac{3}{14x^4y}=\frac{3y^2}{14x^4y^3}\)