Question

Решите пожалуйста B1 с обеих вариантов

Answer 1

$1)\frac{1}{5a^{2}+10ab+20b^{2}} *(a^{3}-8b^{3})=\frac{(a-2b)(a^{2}+2ab+4b^{2})}{5(a^{2}+2ab+4b^{2})}=\frac{a-2b}{5}\\\\a=18;b=-6\\\\\frac{18-2*(-6)}{5}=\frac{18+12}{5}=6$

$2)(a^{3}+27b^{3})*\frac{1}{3a^{2}-9ab+27b^{2}}=\frac{(a+3b)(a^{2}-3ab+9b^{2})}{3(a^{2}-3ab+9b^{2})} =\frac{a+3b}{3}\\\\a=-3;b=5\\\\\frac{-3+3*5}{3}=\frac{-3+15}{3}=4$