Question

( 2-а/2+а - а+2/'а-2) : ( 2+а/2-а + а-2/а+2)​

Answer 1

Ответ:

$\dfrac{4+a^{2}}{4a}$

Пошаговое объяснение:

$\bigg (\dfrac{2-a}{2+a}-\dfrac{a+2}{a-2} \bigg ):\bigg (\dfrac{2+a}{2-a}+\dfrac{a-2}{a+2} \bigg )=\dfrac{(2-a)(a-2)-(a+2)(a+2)}{(a+2)(a-2)}:$

$:\dfrac{(2+a)(2+a)+(2-a)(a-2)}{(2-a)(2+a)}=\dfrac{2a-4-a^{2}+2a-(a^{2}+4a+4)}{a^{2}-4}:$

$:\dfrac{4+4a+a^{2}+2a-4-a^{2}+2a}{4-a^{2}}=\dfrac{4a-4-a^{2}-a^{2}-4a-4}{a^{2}-4}:$

$:\dfrac{8a}{4-a^{2}}=\dfrac{-8-2a^{2}}{a^{2}-4}:\dfrac{8a}{4-a^{2}}=\dfrac{-(8+2a^{2})}{a^{2}-4}:\dfrac{8a}{4-a^{2}}=\dfrac{8+2a^{2}}{-(a^{2}-4)}:\dfrac{8a}{4-a^{2}}=$

$=\dfrac{8+2a^{2}}{4-a^{2}}:\dfrac{8a}{4-a^{2}}=\dfrac{(8+2a^{2})(4-a^{2})}{(4-a^{2})8a}=\dfrac{8+2a^{2}}{8a}=\dfrac{4+a^{2}}{4a};$