Question

Решите, пожалуйста, извините за перевернутое фото

Answer 1

Ответ:

Объяснение:

1)

$\frac{3}{5x^2y} -\frac{2}{3xy^2} =\frac{3*3y-2*5x}{15x^2y^2} =\frac{9y-10x}{15x^2y^2}$

2)

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

3)

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

4)

$\frac{6}{5a-10} -\frac{2}{3a-6} =\frac{6*3-2*5}{3*5(a-2)} =\frac{18-10}{15(a-2)} =\frac{8}{15(a-2)}$

5)

$\frac{1}{x^2-xy} -\frac{1}{xy-y^2} =\frac{y-x}{xy(x-y)} =-\frac{x-y}{xy(x-y)} =-\frac{1}{xy}$

6)

$\frac{x}{2x-2} -\frac{3}{2x+2} +\frac{x-2}{x^2-1} =\frac{x(x+1)-3(x-1)+2(x-2)}{2(x+1)(x-1)} =\\= \frac{x^2+x-3x+3+2x-4}{2(x+1)(x-1)} =\frac{x^2-1}{2(x+1)(x-1)} =\frac{x^2-1}{2(x^2-1)} =\\=\frac{1}{2} =0.5$