Question

Помогите решить пожалуйста ​

Answer 1

Ответ:

1) $\frac{16}{x-4} - \frac{x^2}{x-4} = \frac{16-x^2}{x-4} =\frac{(4-x)*(4+x)}{x-4} = \frac{-(x-4)*(4+x)}{x-4} = (-4+x)=-4-x$

2) $\frac{25}{a+5} - \frac{a^2}{a+5} =\frac{25-a^2}{a+5} = \frac{(5-a)*(5+a)}{a+5} = 5-a$

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

4) $\frac{x-3}{x^-64} + \frac{11}{x^-64} = \frac{x-3+11}{x^2-64} = \frac{x+8}{(x-8)*(x+8)} = \frac{1}{x-8}$

5) $\frac{2a+b}{(a-b)^2} + \frac{2b-5a}{(a-b)^2} = \frac{2a+b+2b-5a}{(a-b)^2} = \frac{-3a+3b}{(a-b)^2} = \frac{-3(a-b)}{(a-b)^2} = \frac{-3}{a-b} = -\frac{3}{a-b}$

6) $\frac{13x+6y}{(x+y)^2} - \frac{11x+4y}{(x+y)^2} = \frac{13x+6y-(11x+4y)}{(x+y)^2} =\frac{13x+6y-11x-4y}{(x+y)^2} = \frac{2x+2y}{(x+y)^2} = \frac{2(x+y)}{(x+y)^2} = \frac{2}{x+y}$