Question

очень РЕШИТЕ ГРУППИРОВКИ! РЕШИТЕ ЗАДАЧИ:Б;В;Г.​

Answer 1

$a)\ \ \dfrac{2x+bx-2y-by}{7x-7y}=\dfrac{2(x-y)+b(x-y)}{7(x-y)}=\dfrac{(x-y)(2+b)}{7(x-y)}=\dfrac{2+b}{7}\\\\\\b)\ \ \dfrac{8a+4b}{2ab+b^2-2ad-bd}=\dfrac{4\, (2a+b)}{b(2a+b)-d(2a+b)}=\dfrac{4\, (2a+b)}{(2a+b)(b-d)}=\dfrac{4}{b-d}$

$c)\ \ \dfrac{xy-x+y-y^2}{x^2-y^2}=\dfrac{y(x-y)-(x-y)}{(x-y)(x+y)}=\dfrac{(x-y)(y-1)}{(x-y)(x+y)}=\dfrac{y-1}{x+y}\\\\\\d)\ \ \dfrac{a^2+2ac+c^2}{a^2+ac-ax-cx}=\dfrac{(a+c)^2}{a(a+c)-x(a+c)}=\dfrac{(a+c)^2}{(a+c)(a-x)}=\dfrac{a+c}{a-x}$

Answer 2

б) $\frac{4(2a+b)}{2a(b-d) +b(b-d)} =\frac{4(2a+b)}{(2a+b)(b-d)} = \frac{4}{(b-d)}$

в) $\frac{y(x-y)-1(x-y)}{(x+y)(x-y)} = \frac{(y-1)(x-y)}{(x+y)(x-y)} = \frac{y-1}{x+y}$

г) $\frac{(a+c)^{2}}{a(a+c)-x(a+c)}=\frac{(a+c)^{2}}{(a-x)(a+c)} = \frac{a+c}{a-x}$