Question

помогите матем даю 10 баллов

Answer 1

$\frac{b}{a+b} * \frac{a^2-b^2}{b^2} = \frac{b*(a-b)(a+b)}{(a+b)*b^2} = \frac{a-b}{b} = \frac{a}{b} - \frac{b}{b} = \frac{a}{b} -1 \\ \\ \frac{2x-2y}{y} : \frac{x^2-y^2}{y^2} = \frac{2(x-y)}{y} * \frac{y^2}{(x-y)(x+y)} = \frac{2(x-y)*y^2}{y(x-y)(x+y)} = \frac{2y}{x+y} \\ \\ 3ab * \frac{a^2}{b} = \frac{3a*a^2 *b}{b} = \frac{3a^3}{1} =3a^3$

$( \frac{a}{a-b} - \frac{a+b}{a} )* \frac{a-b}{2b} = \frac{a*a-(a+b)(a-b)}{a(a-b)} * \frac{a-b}{2b} = \frac{a^2-(a^2-b^2)}{a(a-b)} * \frac{a-b}{2b} = \\ \\ = \frac{a^2-a^2+b^2}{a(a-b)} * \frac{a-b}{2b} = \frac{b^2*(a-b)}{a(a-b)*2b} = \frac{b*1}{a*2} = \frac{b}{2a}$

$\frac{b^2-8b+16}{2b+6} : \frac{b^2-16}{4b+12} = \frac{b^2-2*b*4+4^2}{2(b+3)} * \frac{4(b+3)}{b^2-4^2} = \frac{(b-4)^2 * 4(b+3)}{2(b+3)*(b-4)(b+4)} = \\ \\ =\frac{(b-4) *2*1}{1*1*(b+4)} = \frac{2(b-4)}{b+4} = \frac{2b - 8}{b+4} \\ \\ b=2.4 \\ \\ \frac{2*2.4 - 8}{2.4+4} = \frac{-(8.0-4.8)}{6.4} = \frac{- 3.2}{6.4} =- \frac{32}{64}= - \frac{1}{2}= - 0.5$