ПОМОГИТЕ ПОЖАЛЙУСТА ДАМ 86 БАЛЛОВ !

Question 1

Question 2

Решите задачу:

$1) \frac{2m}{m^2-4} - \frac{2}{m^2-4} :( \frac{m+1}{2m-2} - \frac{2}{m-1} )= \\ \\ =\frac{2m}{(m-2)(m+2)} - \frac{2}{(m-2)(m+2)} :( \frac{m+1}{2(m-1)} - \frac{2}{m-1} )= \\ \\ =\frac{2m}{(m-2)(m+2)} - \frac{2}{(m-2)(m+2)} :( \frac{m+1}{2(m-1)} - \frac{4}{2(m-1)} )= \\ \\=\frac{2m}{(m-2)(m+2)} - \frac{2}{(m-2)(m+2)} : \frac{m+1-4}{2(m-1)}= \\ \\ =\frac{2m}{(m-2)(m+2)} - \frac{2}{(m-2)(m+2)} \cdot \frac{2(m-1)}{m-3}= \\ \\= \frac{2m(m-3)-4(m-1)}{(m-2)(m+2)(m-3)}=$
$=\frac{2m^2-5m+2}{(m-2)(m+2)(m-3)} =\frac{(2m-1)(m-2)}{(m-2)(m+2)(m-3)}=\frac{2m-1}{(m+2)(m-3)}$

$2)( \frac{2}{(a-2)^2} - \frac{9}{4-a^2} ): \frac{4+a^2}{4-a^2} + \frac{2}{a-2} = \\ \\ =( \frac{2}{(a-2)^2} - \frac{a}{(2-a)(2+a)} )\cdot \frac{4-a^2}{4+a^2} + \frac{2}{a-2} = \\ \\ =( \frac{2(2+a)}{(2-a)^2(2+a)} - \frac{a(2-a)}{(2-a)^2(2+a)} )\cdot \frac{4-a^2}{4+a^2} + \frac{2}{a-2} = \\ \\ = \frac{2(2+a)-a(2-a)}{(2-a)^2(2+a)} \cdot \frac{4-a^2}{4+a^2} + \frac{2}{a-2} = \frac{(4+2a-2a+a^2)}{(2-a)} \cdot \frac{1}{(4+a^2)} - \frac{2}{2-a} =$

$= \frac{(4+a^2)}{(2-a)} \cdot \frac{1}{(4+a^2)} - \frac{2}{2-a} =\frac{1}{2-a} - \frac{2}{2-a}=-\frac{1}{2-a}$

$3)( \frac{20x}{25-x^2} + \frac{5-x}{5+x} ): \frac{5+x}{5} - \frac{5}{5-x}=( \frac{20x}{(5-x)(5+x)} + \frac{5-x}{5+x} )\cdot \frac{5}{5+x} - \frac{5}{5-x} = \\ \\ =\frac{20x+(5-x)(5-x)}{(5-x)(5+x)} \cdot \frac{5}{5+x} - \frac{5}{5-x} =\frac{20x+25-10x+x^2}{(5-x)(5+x)} \cdot \frac{5}{5+x} - \frac{5}{5-x} = \\ \\ =\frac{25+10x+x^2}{(5-x)(5+x)} \cdot \frac{5}{5+x} - \frac{5}{5-x} = \frac{5}{5-x} - \frac{5}{5-x} =0$