1: Ta có: \(\frac{\left(a-1\right)^2}{3a+\left(a-1\right)^2}-\frac{1-2a^2+4a}{a^3-1}+\frac{1}{a-1}\)
\(=\frac{\left(a-1\right)^2}{a^2-2a+1+3a}+\frac{2a^2-4a-1}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{1}{a-1}\)
\(=\frac{\left(a-1\right)^2}{a^2+a+1}+\frac{2a^2-4a-1}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{1}{a-1}\)
\(=\frac{\left(a-1\right)^3+2a^2-4a-1+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\frac{a^3-3a^2+3a-1+3a^2-3a}{\left(a-1\right)\left(a^2+a+1\right)}=\frac{a^3-1}{\left(a-1\right)\left(a^2+a+1\right)}=1\)
Ta có: \(M=\left\lbrack\frac{\left(a-1\right)^2}{3a+\left(a-1\right)^2}-\frac{1-2a^2+4a}{a^3-1}+\frac{1}{a-1}\right\rbrack:\frac{a^3+4a}{4a^2}\)
\(=1:\frac{a\left(a^2+4\right)}{4a^2}=\frac{4a^2}{a\left(a^2+4\right)}=\frac{4a}{a^2+4}\)
2: \(a^3+a^2-2a=0\)
=>\(a\left(a^2+a-2\right)=0\)
=>a(a+2)(a-1)=0
=>\(\left[\begin{array}{l}a=0\left(loại\right)\\ a=1\left(loại\right)\\ a=-2\left(nhận\right)\end{array}\right.\)
Thay a=-2 vào M, ta được:
\(M=\frac{4\cdot\left(-2\right)}{\left(-2\right)^2+4}=\frac{-8}{4+4}=\frac{-8}{8}=-1\)
=>M+1=0
