Упр.789 ГДЗ Колягин Ткачёва 8 класс (Алгебра)

1) (a+b)/(a+2b) :(a/(a-2b)+b^2/(a^2-4b^2 ));
2) (b/(b-c)-bc/(b^2-c^2 )) :(4b^2)/(b^2-2bc+c^2 );
3) b^2/(a^2-2ab) :(2ab/(a^2-4b^2 )-b/(a+2b));
4) (2ab/(a^2-9b^2 )-b/(a-3b)) :b^2/(a^2+3ab).

1)

$$ \frac{a+b}{a+2b}:\left(\frac{a}{a-2b}+\frac{b^2}{a^2-4b^2}\right) $$
$$ \frac{a}{a-2b}+\frac{b^2}{a^2-4b^2} = \frac{a(a+2b)+b^2}{(a-2b)(a+2b)} = \frac{(a+b)^2}{(a-2b)(a+2b)} $$
$$ \frac{a+b}{a+2b}:\frac{(a+b)^2}{(a-2b)(a+2b)} = \frac{a+b}{a+2b}\cdot\frac{(a-2b)(a+2b)}{(a+b)^2} = \frac{a-2b}{a+b} $$

2)

$$ \left(\frac{b}{b-c}-\frac{bc}{b^2-c^2}\right):\frac{4b^2}{b^2-2bc+c^2} $$
$$ \frac{b}{b-c}-\frac{bc}{b^2-c^2} = \frac{b(b+c)-bc}{(b-c)(b+c)} = \frac{b^2}{(b-c)(b+c)} $$
$$ \frac{b^2}{(b-c)(b+c)}:\frac{4b^2}{(b-c)^2} = \frac{b^2}{(b-c)(b+c)}\cdot\frac{(b-c)^2}{4b^2} = \frac{b-c}{4(b+c)} $$

3)

$$ \frac{b^2}{a^2-2ab}:\left(\frac{2ab}{a^2-4b^2}-\frac{b}{a+2b}\right) $$
$$ \frac{2ab}{a^2-4b^2}-\frac{b}{a+2b} = \frac{2ab-b(a-2b)}{(a-2b)(a+2b)} = \frac{b(a+2b)}{(a-2b)(a+2b)} = \frac{b}{a-2b} $$
$$ \frac{b^2}{a(a-2b)}:\frac{b}{a-2b} = \frac{b^2}{a(a-2b)}\cdot\frac{a-2b}{b} = \frac{b}{a} $$

4)

$$ \left(\frac{2ab}{a^2-9b^2}-\frac{b}{a-3b}\right):\frac{b^2}{a^2+3ab} $$
$$ \frac{2ab}{a^2-9b^2}-\frac{b}{a-3b} = \frac{2ab-b(a+3b)}{(a-3b)(a+3b)} = \frac{b(a-3b)}{(a-3b)(a+3b)} = \frac{b}{a+3b} $$
$$ \frac{b}{a+3b}:\frac{b^2}{a(a+3b)} = \frac{b}{a+3b}\cdot\frac{a(a+3b)}{b^2} = \frac{a}{b} $$

Ответ: 1) $$\frac{a-2b}{a+b}$$; 2) $$\frac{b-c}{4(b+c)}$$; 3) $$\frac{b}{a}$$; 4) $$\frac{a}{b}$$.