\(ĐKXĐ:\left\{{}\begin{matrix}x\ne0\\x\ne1\end{matrix}\right.\)
a) \(M=\left[\frac{\left(a-1\right)^2}{3a+\left(a-1\right)^2}-\frac{1-2a^2+4a}{a^3-1}+\frac{1}{a-1}\right]:\frac{a^3+4a}{4a^2}\)
\(\Leftrightarrow M=\left[\frac{\left(a-1\right)^2}{3a+a^2-2a+1}-\frac{1-2a^2+4a}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{1}{a-1}\right]:\frac{a^3+4a}{4a^2}\)
\(\Leftrightarrow M=\left[\frac{\left(a-1\right)^2}{a^2+a+1}-\frac{1-2a^2+4a}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{1}{a-1}\right].\frac{4a^2}{a^3+4a}\)
\(\Leftrightarrow M=\frac{\left(a-1\right)^3-\left(1-2a^2+4a\right)+\left(a^2+a+1\right)}{\left(a^2+a+1\right)\left(a-1\right)}.\frac{4a^2}{a^3+4a}\)
\(\Leftrightarrow M=\frac{a^3-3a^2+3a-1-1+2a^2-4a+a^2+a+1}{a^3-1}.\frac{4a^2}{a^3+4a}\)
\(\Leftrightarrow M=\frac{a^3-1}{a^3-1}.\frac{4a^2}{a^3+4a}\)
\(\Leftrightarrow M=\frac{4a^2}{a^3+4a}\)
\(\Leftrightarrow M=\frac{4a}{a^2+4}\)
b) Ta có :
\(\left(a-2\right)^2\ge0\)
\(\Leftrightarrow a^2-4a+4\ge0\)
\(\Leftrightarrow a^2+4\ge4a\)
Dấu " = " xảy ra khi và chỉ khi :
\(\left(a-2\right)^2=0\)
\(\Leftrightarrow a=2\)
Vậy \(Max_M=1\Leftrightarrow a=2\)