Question

ПОМОГИТЕ ПОЖАЛЙУСТА ДАМ 86 БАЛЛОВ !

Answer 1

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

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

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

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