Question

Решите, пожалуйста: 10,11,12,13,14 Если не все, то какие-нибудь Заранее спасибо :3

Answer 1

Решение номеров 10 и 11.............

Answer 2

$(\frac{a^{2}+24}{(2a-5)^{2}}+\frac{8}{5-2a}):(\frac{1}{(2a-5)^{2}}-\frac{2}{2a^{2}+a-15}+\frac{1}{(a+3)^{2}})$

$1) \frac{a^{2}+24}{(2a-5)^{2}}+\frac{8}{5-2a}=\frac{a^{2}+24}{(2a-5)^{2}}-\frac{8}{2a-5}=\frac{a^{2}+24-8(2a-5)}{(2a-5)^{2}}=\frac{a^{2}+24-16a+40}{(2a-5)^{2}}= \\ \\ =\frac{a^{2}-16a+64}{(2a-5)^{2}}=\frac{(a-8)^{2}}{(2a-5)^{2}}$

$2) \frac{1}{(2a-5)^{2}}-\frac{2}{2a^{2}+a-15}+\frac{1}{(a+3)^{2}}=\frac{1}{(2a-5)^{2}}-\frac{2}{(2a-5)(a+3)}+\frac{1}{(a+3)^{2}}= \\ \\ =\frac{(a+3)^{2}-2(2a-5)(a+3)+(2a-5)^{2}}{(2a-5)^{2}(a+3)^{2}}=\frac{a^{2}+6a+9-4a^{2}-2a+30+4a^{2}-20a+25}{(2a-5)^{2}(a+3)^{2}}= \\ \\ =\frac{a^{2}-16a+64}{(2a-5)^{2}(a+3)^{2}}=\frac{(a-8)^{2}}{(2a-5)^{2}(a+3)^{2}}$

$3) \frac{(a-8)^{2}}{(2a-5)^{2}}:\frac{(a-8)^{2}}{(2a-5)^{2}(a+3)^{2}}=\frac{(a-8)^{2}}{(2a-5)^{2}}*\frac{(2a-5)^{2}(a+3)^{2}}{(a-8)^{2}}=(a+3)^{2}$