Question

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Answer 1

Ответ:

$\displaystyle \frac{(a-b)^2}{a}\cdot\left( \frac{a}{a^2-2ab+b^2}+\frac{a}{b^2-a^2}\right)+\frac{3a+b}{a+b}=3$

Объяснение:

$\displaystyle \frac{(a-b)^2}{a}\cdot\left( \frac{a}{a^2-2ab+b^2}+\frac{a}{b^2-a^2}\right)+\frac{3a+b}{a+b}=\\\\\frac{(a-b)^2}{a}\cdot\left( \frac{a}{(a-b)^2}-\frac{a}{a^2-b^2}\right)+\frac{3a+b}{a+b}=\\\\\frac{(a-b)^2}{a}\cdot\left( \frac{a}{(a-b)^2}-\frac{a}{(a-b)(a+b)}\right)+\frac{3a+b}{a+b}=\\\\\frac{(a-b)^2}{a}\cdot\left( \frac{a(a+b)}{((a-b)^2(a+b)}-\frac{a(a-b)}{(a-b)^2(a+b)}\right)+\frac{3a+b}{a+b}=\\\\\frac{(a-b)^2}{a}\cdot\frac{a(a+b)-a(a-b)}{(a-b)^2(a+b)}+\frac{3a+b}{a+b}=$

$\displaystyle \frac{1}{a}\cdot\frac{a(a+b)-a(a-b)}{a+b}+\frac{3a+b}{a+b}=\\\\\frac{1}{a}\cdot\frac{a^2+ab-a^2+ab}{a+b}+\frac{3a+b}{a+b}=\\\\\frac{1}{a}\cdot\frac{2ab}{a+b}+\frac{3a+b}{a+b}=\\\\\frac{2b}{a+b}+\frac{3a+b}{a+b}=\\\\\frac{2b+3a+b}{a+b}=\\\\\frac{3a+3b}{a+b}=\\\\\frac{3(a+b)}{a+b}=\\\\3$