Question

срочнооооооо!срочнооооооо!​

Answer 1

Ответ:

$-1$

Объяснение:

$\displaystyle\\\left( \dfrac{1}{a^2-3a}-\dfrac{1}{3-a}+1\right)\times\left( \dfrac{1}{a^2-1}-\dfrac{1}{(a-1)^3}\right)-\dfrac{a^2}{a^2-1}=\\\\\\=\left( \dfrac{1}{a(a-3)}-\dfrac{1}{3-a}+1\right)\times\left( \dfrac{1}{(a-1)(a+1)}-\dfrac{1}{(a-1)^3}\right)-\dfrac{a^2}{(a-1)(a+1)}=\\\\\\=\left( \dfrac{1}{a(a-3)}+\dfrac{1}{a-3}+1\right)\times\left( \dfrac{1}{(a-1)(a+1)}-\dfrac{1}{(a-1)^3}\right)-\dfrac{a^2}{(a-1)(a+1)}=$

$=\dfrac{1+a+a(a-3)}{a(a-3)}\times \dfrac{(a-1)^2-(a+1)}{(a-1)^3(a+1)}-\dfrac{a^2}{(a-1)(a+1)}=\\\\\\=\dfrac{1+a+a^2-3a}{a(a-3)}\times \dfrac{a^2-2a+1-(a+1)}{(a-1)^3(a+1)}-\dfrac{a^2}{(a-1)(a+1)}=\\\\\\=\dfrac{a^2-2a+1}{a(a-3)}\times \dfrac{a^2-2a+1-a-1}{(a-1)^3(a+1)}-\dfrac{a^2}{(a-1)(a+1)}=\\\\\\=\dfrac{(a-1)^2}{a(a-3)}\times \dfrac{a^2-3a}{(a-1)^3(a+1)}-\dfrac{a^2}{(a-1)(a+1)}=\\\\\\=\dfrac{1}{a(a-3)}\times \dfrac{a(a-3)}{(a-1)(a+1)}-\dfrac{a^2}{(a-1)(a+1)}=$

$=\dfrac{1}{1}\times \dfrac{1}{(a-1)(a+1)}-\dfrac{a^2}{(a-1)(a+1)}=\dfrac{1-a^2}{(a-1)(a+1)}=\\\\\\=-\dfrac{a^2-1}{(a-1)(a+1)}=-\dfrac{(a-1)(a+1)}{(a-1)(a+1)}=\boxed{-1}$