Với `x \ne -1;x \ne 2` có:
`A=2/[x+1]-1/[x-2]-[3x-11]/[x^2-x-2]`
`A=[2(x-2)-x-1-3x+11]/[(x+1)(x-2)]`
`A=[2x-4-x-1-3x+11]/[(x+1)(x-2)]`
`A=[-2x+6]/[(x+1)(x-2)]`
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Với `x \e 3,x \ne 4,x \ne -2` có:
`B=(2/[x-3]-[2x-1]/[x^2-x-6]):5/[x-4]`
`B=(2/[x-3]-[2x-1]/[(x-3)(x+2)]).[x-4]/5`
`B=[2(x+2)-2x+1]/[(x-3)(x+2)].[x-4]/5`
`B=[2x+4-2x+1]/[(x-3)(x+2)].[x-4]/5`
`B=[x-4]/[(x-3)(x+2)]`
\(A=\dfrac{2}{x+1}-\dfrac{1}{x-2}-\dfrac{3x-11}{x^2-x-2}\)
\(=\dfrac{2\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{x+1}{\left(x+1\right)\left(x-2\right)}-\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
\(=\dfrac{-2x+6}{\left(x+1\right)\left(x-2\right)}.\)
\(B=\left(\dfrac{2}{x-3}-\dfrac{2x-1}{x^2-x-6}\right):\dfrac{5}{x-4}\)
\(=\dfrac{5\left(x-4\right)}{\left(x-3\right)\left(x+2\right)}:5=\dfrac{5\left(x-4\right)}{\left(x-3\right)\left(x+2\right)\cdot5}=\dfrac{x-4}{\left(x-3\right)\left(x+2\right)}.\)
