ĐKXĐ: \(\left\{{}\begin{matrix}a-1\ne0\\a^2-1\ne0\\a-a^3\ne0\\a+a^3\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a\ne1\\a\ne\left\{-1;1\right\}\\a\left(1-a^2\right)\ne0\\a\left(1+a^2\right)\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a\ne1\\a\ne\left\{1;-1\right\}\\a\ne\left\{-1;0;1\right\}\\a\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a\ne0\\a\ne-1\\a\ne1\end{matrix}\right.\)
\(M=\frac{a^2}{a-1}+\left(\frac{a}{a^2-1}+\frac{1}{a-a^3}\right):\frac{1-a}{a+a^3}\)
\(=\frac{a^2}{a-1}+\left(\frac{a}{\left(a-1\right)\left(a+1\right)}+\frac{1}{a\left(1-a^2\right)}\right):\frac{1-a}{a\left(1+a^2\right)}\)
\(=\frac{a^2}{a-1}+\left(\frac{a^2}{a\left(a-1\right)\left(a+1\right)}-\frac{1}{a\left(a+1\right)\left(a-1\right)}\right):\frac{1-a}{a\left(1+a^2\right)}\)
\(=\frac{a^2}{a-1}+\frac{\left(a-1\right)\left(a+1\right)}{a\left(a-1\right)\left(a+1\right)}.\frac{a\left(1+a^2\right)}{1-a}\)
\(=\frac{a^2}{a-1}-\frac{1+a^2}{a-1}=\frac{a^2-1-a^2}{a-1}=-\frac{1}{a-1}\)
b/ Thay $a=\frac{1}{2}$ vào M ta được \(M=-\frac{1}{-\frac{1}{2}-1}=-\frac{1}{-\frac{3}{2}}=\frac{1}{\frac{3}{2}}=\frac{2}{3}\)
