Question

с 15 заданием, третий и четвёртый пункт

Answer 1

$3) \ \dfrac{(2x-2y)^{2} }{2x^{2}-2y^{2}}=\dfrac{4(x-y)^{2} }{2(x^{2}-y^{2})} =\dfrac{2(x-y)^{2} }{(x-y)(x+y)} =\dfrac{2(x-y)}{x+y} \\\\x=0,2 \ ; \ y=-0,4\\\\\dfrac{2(x-y)}{x+y}=\dfrac{2(0,2+0,4)}{0,2-0,4}=\dfrac{2\cdot0,6}{-0,2}=\dfrac{1,2}{-0,2}=\boxed{-6}$

$4) \ \dfrac{4x^{2}-40xy+100y^{2}}{15y-3x}=\dfrac{(2x-10y)^{2} }{3(5y-x)}=\dfrac{4(x-5y)^{2} }{3(5y-x)}=\dfrac{4(5y-x)^{2} }{3(5y-x)}=\\\\=\dfrac{4(5y-x)}{3} \\\\x-5y=0,6 \ \Rightarrow \ 5y-x=-0,6\\\\\dfrac{4(5y-x)}{3}=\dfrac{4\cdot(-0,6)}{3}=\dfrac{-2,4}{3} =\boxed{-0,8}$