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Đặt: \(L=\dfrac{3\left(a+2\right)}{a^3+a^2+a+1}+\dfrac{2a^2-a-10}{a^3-a^2+a-1}\)
Ta có:
\(\dfrac{3\left(a+2\right)}{a^3+a^2+a+1}=\dfrac{3\left(a+2\right)}{a^2\left(a+1\right)+1\left(a+1\right)}=\dfrac{3\left(a+2\right)}{\left(a^2+1\right)\left(a+1\right)}\)
\(\dfrac{2a^2-a-10}{a^3-a^2+a-1}=\dfrac{a\left(2a-1\right)-10}{a^2\left(a-1\right)+1\left(a-1\right)}=\dfrac{a\left(2a-1\right)-10}{\left(a^2+1\right)\left(a-1\right)}\)
Như vậy \(L=\dfrac{3\left(a+2\right)}{\left(a^2+1\right)\left(a+1\right)}+\dfrac{a\left(2a-1\right)-10}{\left(a^2+1\right)\left(a-1\right)}\)
Đặt:
\(N=\dfrac{5}{a^2+1}+\dfrac{3}{2a+2}-\dfrac{3}{2a-2}\)
\(N=\dfrac{5}{a^2+1}+\dfrac{3\left(2a-2\right)}{\left(2a+2\right)\left(2a-2\right)}-\dfrac{3\left(2a+2\right)}{\left(2a+2\right)\left(2a-2\right)}\)
\(N=\dfrac{5}{a^2+1}+\dfrac{6a-6}{4a^2-4}-\dfrac{6a+6}{4a^2-4}\)
\(N=\dfrac{5}{a^2+1}+\dfrac{6a-6-6a-6}{4a^2-4}=\dfrac{5}{a^2+1}+\dfrac{-12}{4a^2-4}\)
\(N=\dfrac{5}{a^2+1}+\dfrac{-12}{4\left(a^2-1\right)}=\dfrac{5}{a^2+1}+\dfrac{-3}{a^2-1}\)
\(N=\dfrac{5\left(a^2-1\right)}{\left(a^2+1\right)\left(a^2-1\right)}+\dfrac{-3\left(a^2+1\right)}{\left(a^2-1\right)\left(a^2+1\right)}\)
\(N=\dfrac{5a^2-5-3a^2-3}{a^4-1}=\dfrac{2a^2-8}{a^4-1}\)
Thay M với N vào A Mình cạn sức rồi 
Đặt \(\left(a-1\right)^2=t\)
Ta có: \(\left(a-1\right)^4-11\left(a-1\right)^2+30\)
\(=t^2-11t+30\)
\(=t\left(t-5\right)-6\left(t-5\right)=\left(t-5\right)\left(t-6\right)\)
\(=\left[\left(a-1\right)^2-5\right]\left[\left(a-1\right)^2-6\right]\)
\(=\left(a^2-2a-4\right)\left(a^2-2a-5\right)\)
Đặt \(a^2-2a=k\)
Ta có: \(3\left(a-1\right)^4-18\left(a^2-2a\right)-3\)
\(=3\left(a^2-2a+1\right)^2-18\left(a^2-2a\right)-3\)
\(=3\left(k+1\right)^2-18k-3\)
\(=3k^2+6k+3-18k-3\)
\(=3k^2-12k=3k\left(k-4\right)\)
\(=3\left(a^2-2a\right)\left(a^2-2a-4\right)\)(Ở đây bạn ghi thêm điều kiện nhé)
Khi đó: \(N=\frac{\left(a^2-2a-4\right)\left(a^2-2a-5\right)}{3\left(a^2-2a\right)\left(a^2-2a-4\right)}=\frac{a^2-2a-5}{3\left(a^2-2a\right)}\)
