b, đk x khác -4 ; -5 ; -6 ; -7
\(=\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-...+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\dfrac{3}{\left(x+4\right)\left(x+7\right)}\)
a, đk x khác 3/2 ; -3/2 ; 0
\(\dfrac{7}{8x^2-18}+\dfrac{1}{x\left(2x+3\right)}-\dfrac{1}{2\left(2x-3\right)}=\dfrac{7}{8x^2-18}+\dfrac{4x-6-2x^2-3x}{2x\left(4x^2-9\right)}\)
\(=\dfrac{7}{2\left(4x^2-9\right)}+\dfrac{-2x^2+x-6}{2x\left(4x^2-9\right)}=\dfrac{-2x^2+8x-6}{2x\left(4x^2-9\right)}=\dfrac{-\left(x+1\right)\left(x+3\right)}{2x\left(4x^2-9\right)}\)
