Question

СРОЧНО, АЛГЕБРА ДОВЕДІТЬ ТОТОЖНІСТЬ 8 КЛАСС
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Answer 1

Ответ:

$\displaystyle \left(\frac{2i}{i+4}-\frac{6i}{i^2+8i+16}\right):\frac{i+1}{i^2-16}-\frac{i^2-12i}{i+4}=i$

Объяснение:

$\displaystyle \left(\frac{2i}{i+4}-\frac{6i}{i^2+8i+16}\right):\frac{i+1}{i^2-16}-\frac{i^2-12i}{i+4}=\\\\\\\left(\frac{2i}{i+4}-\frac{6i}{(i+4)^2}\right)\cdot\frac{i^2-16}{i+1}-\frac{i(i-12)}{i+4}=\\\\\\\left(\frac{2i(i+4)}{(i+4)^2}-\frac{6i}{(i+4)^2}\right)\cdot\frac{(i-4)(i+4)}{i+1}-\frac{i(i-12)}{i+4}=\\\\\\\frac{2i(i+4)-6i}{(i+4)^2}\cdot\frac{(i-4)(i+4)}{i+1}-\frac{i(i-12)}{i+4}=$

$\displaystyle\frac{2i^2+8i-6i}{(i+4)^2}\cdot\frac{(i-4)(i+4)}{i+1}-\frac{i(i-12)}{i+4}=\\\\\\\frac{2i^2+2i}{i+4}\cdot\frac{i-4}{i+1}-\frac{i(i-12)}{i+4}=\\\\\\\frac{2i(i+1)}{i+4}\cdot\frac{i-4)}{i+1}-\frac{i(i-12)}{i+4}=\\\\\\\frac{2i(i-4)}{i+4}-\frac{i(i-12)}{i+4}=\frac{2i(i-4)-i(i-12)}{i+4}=$

$\displaystyle\frac{2i^2-8i-i^2+12i}{i+4}=\frac{i^2+4i}{i+4}=\frac{i(i+4)}{i+4}=i$

Answer 2

Объяснение:

$\left( \rule{0pt}{1em} \dfrac{2i}{i+4}- \dfrac{6i}{i^2+8i+16}\right) \div \dfrac{i+1}{i^2-16}-\dfrac{i^2-12i}{i+4}=i\\\\\\\left( \rule{0pt}{1em} \dfrac{2i}{i+4}- \dfrac{6i}{(i+4)^2}\right) \times \dfrac{i^2-16}{i+1}-\dfrac{i(i-12)}{i+4}=i\\\\\\\left( \rule{0pt}{1em} \dfrac{2i(i+4)-6i}{(i+4)^2}\right) \times \dfrac{(i-4)(i+4)}{i+1}-\dfrac{i(i-12)}{i+4}=i\\\\\\ \dfrac{2i^2+8i-6i}{(i+4)^2}\times \dfrac{(i-4)(i+4)}{i+1}-\dfrac{i(i-12)}{i+4}=i$

$\dfrac{2i(i+1)}{i+4}\times \dfrac{i-4}{i+1}-\dfrac{i(i-12)}{i+4}=i\\\\\\ \dfrac{2i}{i+4}\times \dfrac{i-4}{1}-\dfrac{i(i-12)}{i+4}=i\\\\ \\\dfrac{2i(i-4)}{i+4}-\dfrac{i(i-12)}{i+4}=i\\\\\\\dfrac{2i(i-4)-i(i-12)}{i+4}=i\\\\\\\dfrac{2i^2-8i-i^2+12i}{i+4}=i\\\\\\\dfrac{i^2+4i}{i+4}=i\\\\\\\dfrac{i(i+4)}{(i+4)}=i\\\\\\i=i$

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