Question

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Answer 1

Ответ: -1.

Объяснение:

$\displaystyle\\1)\\\\\frac{1}{a^2-3a}-\frac{1}{3-a} +1=\frac{1}{a*(a-3)}+\frac{1}{a-3} +1=\frac{1+a+a^2-3a}{a*(a-3)} =\\\\\\=\frac{a^2-2a+1}{a*(a-3)} =\frac{(a-1)^2}{a^2-3a}.\\ 2)\\\\\frac{1}{a^2-1}-\frac{1}{(a-1)^3}=\frac{1}{(a-1)*(a+1)}-\frac{1}{(a-1)^3} =\frac{(a-1)^2-(a+1)}{(a+1)*(a-1)^3}=\\\\\\ =\frac{a^2-2a+1-a-1}{(a+1)*(a-1)^2}=\frac{a^2-3a}{(a+1)*(a-1)^3} .$

$\displaystyle\\ 3)\\\\ \frac{(a-1)^2}{a^2-3a}*\frac{a^2-3a}{(a+1)*(a-1)^3} =\frac{1}{(a+1)*(a-1)}=\frac{1}{a^2-1} .\\ 4) \\\\\frac{1}{a^2-1} -\frac{a^2}{a^2-1}=\frac{1-a^2}{a^2-1} =-\frac{a^2-1}{a^2-1} =-1.$