Ответ:

[tex]1)\ -2\\2)\ 1[/tex]

Объяснение:

[tex]1)\\\\\frac{3y+7}{4-y}+\frac{y+15}{y-4}=-\frac{3y+7}{y-4}+\frac{y+15}{y-4}=\\\\\frac{-3y-7+y+15)}{y-4}=\frac{-2y+8}{y-4}=\frac{-2(y-4)}{y-4}=-2\\\\2)\\\\\frac{25-3x}{(x-5)^2}-\frac{7x-x^2}{(5-x)^2}=\frac{25-3x}{(x-5)^2}-\frac{7x-x^2}{(x-5)^2}=\\\\\frac{25-3x-(7x-x^2)}{(x-5)^2}=\frac{25-3x-7x+x^2}{(x-5)^2}=\frac{x^2-10x+25}{(x-5)^2}=\frac{(x-5)^2}{(x-5)^2}=1[/tex]