Question

помогите сделать алгебру.Зарание спасибо.номер 290 и 291.

Answer 1

$1) \frac{4}{a+2}- \frac{3}{a-2}+ \frac{12}{ a^{2}-4 }= \frac{4a-8-3a-6+12}{(a+2)(a-2)}= \frac{1}{a+2}$
$2) \frac{6m}{m-3}+ \frac{m^2+m}{4}* \frac{8}{m+1}= \frac{2m^2+2m}{m+1} +\frac{6m}{m-3}= \frac{(2m^2+2m)(m-3+6m(m+1))}{(m-3)(m+1)}=$
$=2m^2+8m$
$3) \frac{x}{x-y}- \frac{x}{x+y} ): \frac{xy}{x+y} = \frac{x(x+y-x(x-y))}{(x-y)(x+y)}= \frac{2xy}{(y-2)(x+y)}* \frac{x+y}{xy} = \frac{xy}{x-y}$
$4) \frac{y}{2x+2y}+ \frac{x}{a-1}* \frac{a-1}{2x+2y}= \frac{x+y}{2x+2y}= \frac{1}{2}=0.5$
№291
$1)( \frac{x}{x+1} +1)* \frac{x+1}{2x-1}= \frac{2x+1}{x+1}* \frac{x+1}{2x-1}\ =\frac{2x+1}{2x-1}$
$2) (\frac{4y}{2-y}- y): \frac{y-2}{y+2} = \frac{2y+y^2}{2-y}* \frac{y-2}{y+2}= \frac{y(2+y)}{y+2}=y$
$3) \frac{a+3}{a^2+9}*( \frac{a+3}{a-3}+ \frac{a-3}{a+3})= \frac{a+3}{a^2+9}* \frac{2a^2}{(a-3)(a+3)}= \frac{2a^2}{(a-3)(a^2+9)}$