Упр.6.9 ГДЗ Мордкович 8 класс (Алгебра)

6.9

1. а)
   $$\frac{1}{x+y}+\frac{1}{x-y}=\frac{x-y+x+y}{(x+y)(x-y)}=\frac{2x}{x^2-y^2}.$$
   Тогда
   $$\frac{\frac{1}{x+y}+\frac{1}{x-y}}{\frac{1}{x+y}-\frac{1}{x-y}}=\frac{\frac{2x}{x^2-y^2}}{\frac{x-y-x-y}{(x+y)(x-y)}}=\frac{\frac{2x}{x^2-y^2}}{\frac{-2y}{x^2-y^2}}=-\frac{x}{y}.$$

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

Ответ

а) $$-\frac{x}{y}$$

б) $$\frac{1}{4}$$