a)ĐKXĐ : a khác -1,1
\(C=\dfrac{4a+1}{5a^2-5}+\dfrac{1}{5a+5}+\dfrac{1}{1-a}=\dfrac{4a+1}{5a^2-5}+\dfrac{1-a+5a+5}{\left(5a+5\right)\left(1-a\right)}=\dfrac{4a+1}{5a^2-5}+\dfrac{4a+6}{5-5a^2}=\dfrac{4a+1-4a-6}{5a^2-5}=\dfrac{-5}{5a^2-5}=\dfrac{-1}{a^2-1}\)
Vậy...