a: \(M=\left[\dfrac{\left(a-1\right)^2}{3a+\left(a-1\right)^2}-\dfrac{1-2a^2+4a}{a^3-1}\right]:\dfrac{a^3+4a}{4a^2}\)
\(=\left[\dfrac{\left(a-1\right)^2}{a^2+a+1}+\dfrac{2a^2-4a-1}{\left(a-1\right)\left(a^2+a+1\right)}\right]\cdot\dfrac{4a^2}{a^3+4a}\)
\(=\dfrac{\left(a-1\right)^3+\left(2a^2-4a-1\right)}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{4a}{a^2+4}\)
\(=\dfrac{a^3-3a^2+3a-1+2a^2-4a-1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{4a}{a^2+4}\)
\(=\dfrac{a^3-a^2-a-2}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{4a}{a^2+4}\)
\(=\dfrac{a^3-2a^2+a^2-2a+a-2}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{4a}{a^2+4}=\dfrac{\left(a-2\right)\left(a^2+a+1\right)}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{4a}{a^2+4}\)
\(=\dfrac{a-2}{a-1}\cdot\dfrac{4a}{a^2+4}=\dfrac{4a\left(a-2\right)}{\left(a-1\right)\left(a^2+4\right)}\)
b: Để M>0 thì \(\dfrac{4a\left(a-2\right)}{\left(a-1\right)\left(a^2+4\right)}>0\)
=>\(\dfrac{a\left(a-2\right)}{a-1}>0\)
TH1: \(\left\{{}\begin{matrix}a\left(a-2\right)>0\\a-1>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}a>2\\a< 0\end{matrix}\right.\\a>1\end{matrix}\right.\)
=>Loại
TH2: \(\left\{{}\begin{matrix}a\left(a-2\right)< 0\\a-1< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}0< a< 2\\a< 1\end{matrix}\right.\)
=>0<a<1
