Страница 33 номер 1.99, ГДЗ по алгебре за 8 класс к учебнику Дорофеева

Преобразование каждого выражения выполняем по действиям.

а)

1) \(\frac{x}{x+1}+\frac{x^2+1}{1-x^2}-\frac{x}{x-1} = \frac{x(x-1)-(x^2+1)-x(x+1)}{(x-1)(x+1)} = \frac{x^2-x-x^2-1-x^2-x}{(x-1)(x+1)} = \frac{-(x+1)^2}{(x-1)(x+1)} = \frac{x+1}{1-x}\);

2) \(\frac{x+1}{1-x}:\frac{x+x^2}{(1-x)^2} = \frac{(x+1)(1-x)^2}{(1-x)x(1+x)} = \frac{1-x}{x}\).

б)

1) \(\frac{5}{a+5}-\frac{a^2+25}{a^2-25}-\frac{5}{5-a} = \frac{5(a-5)-(a^2+25)+5(a+5)}{(a-5)(a+5)} = \frac{5a-25-a^2-25+5a+25}{(a-5)(a+5)} = \frac{-(a-5)^2}{(a-5)(a+5)} = \frac{5-a}{a+5}\);

2) \(\frac{(a-5)^2}{a^2+5a}:\frac{5-a}{a+5} = \frac{(a-5)^2(a+5)}{a(a+5)(5-a)} = \frac{(5-a)^2(a+5)}{a(a+5)(5-a)} = \frac{5-a}{a}\).

в)

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

2) \(\frac{x+y}{xy(x-y)}:\frac{x^2+y^2+2xy}{x^2y-xy^2} = \frac{(x+y)xy(x-y)}{xy(x-y)(x+y)^2} = \frac{1}{x+y}\);

3) \(\frac{1}{x+y}-\frac{x}{x^2-y^2} = \frac{x-y-x}{(x+y)(x-y)} = \frac{-y}{x^2-y^2} = \frac{y}{y^2-x^2}\).

г)

1) \(\frac{2}{2b+b^2}-\frac{b}{4+2b} = \frac{2}{b(2+b)}-\frac{b}{2(2+b)} = \frac{4-b^2}{2b(2+b)} = \frac{(2-b)(2+b)}{2b(2+b)} = \frac{2-b}{2b}\);

2) \(\frac{4b}{4-4b+b^2}\cdot\frac{2-b}{2b} = \frac{4b(2-b)}{(b-2)^2 \cdot 2b} = \frac{-2}{b-2} = \frac{2}{2-b}\);

3) \(\frac{4}{b^2-4}+\frac{2}{2-b} = \frac{4-2(b+2)}{(b-2)(b+2)} = \frac{-2b}{b^2-4} = \frac{2b}{4-b^2}\).

Ответ: а) \(\frac{1-x}{x}\); б) \(\frac{5-a}{a}\); в) \(\frac{y}{y^2-x^2}\); г) \(\frac{2b}{4-b^2}\).