Question

решите пж даю 50 балов

Answer 1

$\frac{9-4y^2}{4y^2-6y+9}:\frac{3-2y}{8y^3+27} = \frac{9-4y^2}{4y^2-6y+9}*\frac{8y^3+27}{3-2y} =\frac{(3-2y)(3+2y)}{4y^2-6y+9}*\frac{(2y+3)(4y^2-6y+9)}{3-2y} =$$=\frac{3+2y}{1}*\frac{(2y+3)*1}{1} = (3+2y)^2$

если $y=-\frac{1}{2},$   то   $(3+2y)^2=(3+2*(-\frac{1}{2}))^2=(3-1)^2=4$

P.S.

$a^2-b^2=(a-b)(a+b)$

$a^3+b^3=(a+b)(a^2-ab+b^2)$