Question

24 номер, пожалуйста.

Answer 1

$( \frac{2a+10}{3a-1}- \frac{130-a}{3a-1}+ \frac{30-3a}{a} )* \frac{4a(3a^2+8a-3)}{4-a^2}=$$( \frac{3a-120}{3a-1}+ \frac{30-3a}{a})* \frac{4a(3(a- \frac{1}{3} )(a+3))}{4-a^2}=$$\frac{3a^2-120a+120a-9a^2-30+3a}{a(3a-1)} * \frac{4a(3a-1)(a+3)}{4-a^2}= \frac{6(a-2)(a+2.5)(4a+12)}{(a-2)(a+2)} =$$\frac{12(2a+5)(a+3)}{a+2}$