Монгол Бодлогын Сан

Заавар: $$a^2+2ab+b^2=(a+b)^2\Rightarrow 1+2y+y^2=(1+y)^2$$ $$a^3-3a^2b+3ab^2-b^3=(a-b)^3\Rightarrow 1-3y+3y^2-y^3=(1-y)^3$$ $$a^2-b^2=(a-b)(a+b)\Rightarrow y^2-1=(y-1)(y+1)$$

Бодолт: \begin{align*} \text{Илэрх.}&=\dfrac{1+y^2}{1+2y+y^2}+\dfrac{1-3y+3y^2-y^3}{y^2-1}\cdot\dfrac{1}{1+y}\\ &=\dfrac{1+y^2}{(1+y)^2}+\dfrac{(1-y)^3}{(y-1)(y+1)}\cdot\dfrac{1}{1+y}\\ &=\dfrac{1+y^2}{(1+y)^2}+\dfrac{-\cancel{(y-1)}(1-y)^2}{\cancel{(y-1)}(y+1)^2}\\ &=\dfrac{1+y^2}{(1+y)^2}-\dfrac{(1-y)^2}{(y+1)^2}=\dfrac{1+y^2-(1-2y+y^2)}{(1+y)^2}\\ &=\dfrac{2y}{(1+y)^2} \end{align*}