Question

помогите по братски

Answer 1

1) $\frac{1}{2-a}- \frac{1}{2+a}+ \frac{a}{a^{2}-4 }- \frac{a^{2}+4 }{8a-2a^{3} } =\frac{1}{2-a}- \frac{1}{2+a}- \frac{a}{4-a^{2} }- \frac{a^{2}+4 }{2a*(4-a^{2}) } =$

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

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

2) $\frac{2x-5}{x^{2}-6x+9 } -\frac{x+8}{(3-x)^{2}}+\frac{10}{9+x^{2}-6x } =\frac{2x-5}{(x-3)^{2}} - \frac{x+8}{(x-3)^{2}}+\frac{10}{(x-3)^{2} } =$$=\frac{2x-5-x-8+10}{(x-3)^{2}} =\frac{x-3}{(x-3)^{2}}=\frac{1}{x-3}$

Подставляем значение x = 3,1

1 : (3,1-3) = 1 : 0,1 = 10