Question

ДАЮ 50Б
Доведіть тотожність : на картинці ❗️

Answer 1

Ответ:

$\displaystyle1)\ \left(\frac{ab}{a^2-b^2}+\frac{b}{2b-2a}\right):\frac{2b}{a^2-b^2}=\frac{a-b}{4}$

$\displaystyle2) \ \left(\frac{8a}{4-a^2}-\frac{a-2}{a+2}\right):\frac{a+2}{a}+\frac{2}{a-2}=-1$

Объяснение:

$\displaystyle1)\\\\\left(\frac{ab}{a^2-b^2}+\frac{b}{2b-2a}\right):\frac{2b}{a^2-b^2}=\\\\\\\left(\frac{ab}{(a-b)(a+b)}-\frac{b}{2a-2b}\right):\frac{2b}{(a-b)(a+b)}=\\\\\\\left(\frac{ab}{(a-b)(a+b)}-\frac{b}{2(a-b)}\right)\cdot \frac{(a-b)(a+b)}{2b}=\\\\\\\left(\frac{2ab}{2(a-b)(a+b)}-\frac{b(a+b)}{2(a-b)(a+b)}\right)\cdot \frac{(a-b)(a+b)}{2b}=$

$\displaystyle \\\\\\\frac{2ab-b(a+b)}{2(a-b)(a+b)}\cdot \frac{(a-b)(a+b)}{2b}=\\\\\\\frac{2ab-ab-b^2}{2}\cdot \frac{1}{2b}=\frac{ab-b^2}{4b}=\frac{b(a-b)}{4b}=\frac{a-b}{4}$

$2)\\\\\displaystyle\left(\frac{8a}{4-a^2}-\frac{a-2}{a+2}\right):\frac{a+2}{a}+\frac{2}{a-2}=\\\\\\\left(\frac{-8a}{a^2-4}-\frac{a-2}{a+2}\right)\cdot \frac{a}{a+2}+\frac{2}{a-2}=\\\\\left(\frac{-8a}{(a-2)(a+2)}-\frac{a-2}{a+2}\right)\cdot \frac{a}{a+2}+\frac{2}{a-2}=\\\\\\\left(\frac{-8a}{(a-2)(a+2)}-\frac{(a-2)^2}{(a-2)(a+2)}\right)\cdot \frac{a}{a+2}+\frac{2}{a-2}=\\\\\\\frac{-8a-(a-2)^2}{(a-2)(a+2)}\cdot \frac{a}{a+2}+\frac{2}{a-2}=$

$\displaystyle\frac{-8a-(a^2-4a+4)}{(a-2)(a+2)}\cdot \frac{a}{a+2}+\frac{2}{a-2}=\\\\\\\frac{-8a-a^2+4a-4}{(a-2)(a+2)}\cdot \frac{a}{a+2}+\frac{2}{a-2}=\\\\\\\frac{-a^2-4a-4}{(a-2)(a+2)}\cdot \frac{a}{a+2}+\frac{2}{a-2}=\\\\\\\frac{-(a^2+4a+4)}{(a-2)(a+2)}\cdot \frac{a}{a+2}+\frac{2}{a-2}=\\\\\\\frac{-(a+2)^2}{(a-2)(a+2)}\cdot \frac{a}{a+2}+\frac{2}{a-2}=\\\\\\\frac{-1}{a-2}\cdot \frac{a}{1}+\frac{2}{a-2}=\\\\\\\frac{-a}{a-2}+\frac{2}{a-2}=\frac{-a+2}{a-2}=\frac{-(a-2)}{a-2}=-1$

Answer 2

Відповідь:

Пояснення:

доведення.