\(A=\left(\frac{1-a^3}{a-a^2}+1\right)\cdot\left(\frac{1+a^3}{1+a}-a\right):\frac{\left(1-a^2\right)^3}{1+a}\)
\(=\left(\frac{\left(1-a\right)\cdot\left(1+a+a^2\right)}{a\cdot\left(1-a\right)}+1\right)\cdot\left(\frac{\left(1+a\right)\left(1-a+a^2\right)}{1+a}-a\right)\)\(:\frac{\left(1-a\right)^3\cdot\left(1+a\right)^3}{1+a}\)
\(=\left(\frac{1+a+a^2+a}{a}\right)\cdot\left(1-a+a^2-a\right):\left[\left(1-a\right)^3\cdot\left(1+a\right)^2\right]\)
\(=\frac{1+2a+a^2}{a}\cdot\left(1-2a+a^2\right):\left[\left(1-a\right)^3\cdot\left(1+a\right)^2\right]\)
\(=\frac{\left(1+a\right)^2}{a}\cdot\left(1-a\right)^2:\left[\left(1-a\right)^3\cdot\left(1+a\right)^2\right]\)
\(=\text{[}\frac{\left(1+a\right)^2}{a}:\left(1+a\right)^2\text{]}\cdot\text{[}\left(1-a\right)^2:\left(1-a\right)^3\text{]}\)
\(=\frac{1}{a}\cdot\frac{1}{1-a}=\frac{1}{a\left(1-a\right)}=\frac{1}{a-a^2}\)
Để \(A>A^2\Rightarrow\frac{1}{a-a^2}>\frac{1}{\left(a-a^2\right)^2}\)
Có ĐKXĐ : \(\left(a-a^2\right)\ne0\)
Mà \(\left(a-a^2\right)< \left(a-a^2\right)^2\)trừ trường hợp \(\left(a-a^2\right)=1\)
Từ tất cả điều trên suy ra : \(A\)thuộc tất cả các giá trị khác 1 để \(A>A^2\)
