Question

Помогите пожалуйста.

Answer 1

$3(a-2)^2-6a(a-2)=3(a^2-4a+4)-6a^2+12a= \\ =3a^2-12a+12-6a^2+12a=-3a^2+12=12-3*( \sqrt{2} )^2= \\ =12-3*2=12-6=6 \\ \\ ( \frac{y}{x} - \frac{x}{y} ): \frac{2x-2y}{xy^3}= \frac{y^2-x^2}{xy} * \frac{xy^3}{2(x-y)} = \frac{(y-x)(y+x)y^2}{2(x-y)}=- \frac{1}{2}(x+y)y^2= \\ = - \frac{1}{2}(4- \sqrt{5} + \sqrt{5} )( \sqrt{5} )^2=- \frac{1}{2}*4*5=-10$

Answer 2

$3(a^2-4a+4)-6a^2+12a=3a^2-12a+12-6a^2+12a=-3a^2+12;$ $-3*( \sqrt{2})^2+12=-3*2+12=6$
$\frac{y^2- x^{2} }{xy} * \frac{xy^3}{2(x-y)}= -\frac{(x-y)(x+y)xy^3}{2xy(x-y)}= -\frac{(x+y)y^2}{2}= -\frac{(4- \sqrt{5}+ \sqrt{5} )*( \sqrt{5})^2 }{2}=- \frac{4*5}{2}=-10$