Монгол Бодлогын Сан

Заавар:$$a^2-2a+1=(a-1)^2$$ $$a^3-1=(a-1)(a^2+a+1)$$

Бодолт:\begin{align*} \text{Илэрх.}&=\left(\dfrac{a^2}{a^2-2a+1}-\dfrac{a^2+a}{a^3-1}\cdot\Bigl(\dfrac{1}{a^2-a}^{\color{red}{(a+1}}+\dfrac{a}{a^2-1}^{\color{red}{(a}}\biggr)\right):\dfrac{a}{a-1}\\ &=\left(\dfrac{a^2}{(a-1)^2}-\dfrac{a(a+1)}{(a-1)(a^2+a+1)}\cdot\dfrac{a^2+a+1}{a(a-1)(a+1)}\right):\dfrac{a}{a-1}\\ &=\left(\dfrac{a^2}{(a-1)^2}-\dfrac{1}{(a-1)^2}\right):\dfrac{a}{a-1}\\ &=\dfrac{a^2-1}{(a-1)^2}:\dfrac{a}{a-1}=\dfrac{(a-1)(a+1)}{(a-1)^2}:\dfrac{a}{a-1}\\ &=\dfrac{a+1}{a-1}:\dfrac{a}{a-1}=\dfrac{a+1}{a-1}\cdot\dfrac{a-1}{a}=\dfrac{a+1}{a} \end{align*}