Question

Выражение: 1) 2) \frac{4-9c^2}{9c^2-12c+16} : \frac{2-3c}{27c^3+64}

Answer 1

$\frac{64a^3-27b^3}{(4a-3b)^2} \cdot \frac{9b^2-16a^2}{16a^2+12ab+9b^2}= \frac{(4a-3b)(16a^2+12ab+9b^2)}{(4a-3b)^2}\cdot \frac{(3b-4a)(3b+4a)}{16a^2+12ab+9b^2}=\\\\ = \frac{(3b-4a)(3b+4a)}{4a-3b}=- \frac{(4a-3b)(4a+3b)}{4a-3b} =-(4a+3b)=-4a-3b$

$\frac{4-9c^2}{9c^2-12c+16}: \frac{2-3c}{27c^3+64}= \frac{(2-3c)(2+3c)}{9c^2-12c+16}\cdot \frac{(3c+4)(9c^2-12c+16)}{2-3c}=\\\\=(2+3c)(3c+4)=9c^2+18c+8$