Решите пожалуйста........

1) $( \frac{x}{x+1} + \frac{1}{1} )* \frac{1+x}{2x-1} =( \frac{x}{x+1} + \frac{x+1}{x+1} )*\frac{1+x}{2x-1}= \frac{x+x+1}{x+1} *\frac{1+x}{2x-1}=$
$= \frac{2x+1}{x+1} *\frac{x+1}{2x-1}= \frac{2x+1}{2x-1}$

2) $( \frac{2m+1}{2m-1} - \frac{2m-1}{2m+1} ): \frac{4m}{10m-5} =( \frac{(2m+1)^{2} }{(2m-1)(2m+1)} - \frac{(2m-1)^{2} }{(2m-1)(2m+1)} )* \frac{5(2m-1)}{4m} =$
$= \frac{(2m+1)^{2}-(2m-1)^{2} }{(2m-1)(2m+1)} * \frac{5(2m-1)}{4m} = \frac{5(2m+1+2m-1)(2m+1-2m+1)}{4m(2m+1)} =$
$= \frac{5*4m*2}{4m(2m+1)} = \frac{10}{2m+1}$

3) $\frac{ a^{2}-25 }{a+3} * \frac{1}{ a^{2}+5a } - \frac{a+5}{a^{2}-3a } = \frac{(a-5)(a+5)}{a+3} * \frac{1}{a(a+5)} - \frac{a+5}{a(a-3)} =$
$= \frac{a-5}{a(a+3)} - \frac{a+5}{a(a-3)} = \frac{(a-5)(a-3)}{a(a+3)(a-3)} - \frac{(a+5)(a+3)}{a(a+3)(a-3)} =$
$=\frac{(a-5)(a-3)-(a+5)(a+3)}{a(a-3)(a+3)} = \frac{ a^{2}-5a-3a+15- a^{2}-5a-3a-15 }{a(a-3)(a+3)} = \frac{-16a}{a( a^{2}-9) } =- \frac{16}{ a^{2}-9 }$