Выполните действия:
1) $\frac{a + 2}{a^2 - 2a + 1} : \frac{a^2 - 4}{3a - 3} - \frac{3}{a - 2}$;
2) $\frac{b^2 + 3b}{b^3 + 9b} * (\frac{b - 3}{b + 3} + \frac{b + 3}{b - 3})$;
3) $(\frac{3c + 1}{3c - 1} - \frac{3c - 1}{3c + 1}) : \frac{2c}{6c + 2}$;
4) $(\frac{1}{a^2 - 4ab + 4b^2} - \frac{1}{4b^2 - a^2}) : \frac{2a}{a^2 - 4b^2}$;
5) $(\frac{a - 8}{a^2 - 10a + 25} - \frac{a}{a^2 - 25}) : \frac{a - 20}{(a - 5)^2}$;
6) $(\frac{2x + 1}{x^2 + 6x + 9} - \frac{x - 2}{x^2 + 3x}) : \frac{x^2 + 6}{x^3 - 9x}$.

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Решение

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ГДЗ учебник по алгебре 8 класс Мерзляк, Полонский, Якир. §6. Упражнения. Номер №178

Решение 1

$\frac{a + 2}{a^2 - 2a + 1} : \frac{a^2 - 4}{3a - 3} - \frac{3}{a - 2} = \frac{a + 2}{(a - 1)^2} : \frac{(a - 2)(a + 2)}{3(a - 1)} - \frac{3}{a - 2} = \frac{a + 2}{(a - 1)^2} * \frac{3(a - 1)}{(a - 2)(a + 2)} - \frac{3}{a - 2} = \frac{1}{a - 1} * \frac{3}{a - 2} - \frac{3}{a - 2} = \frac{3}{(a - 1)(a - 2)} - \frac{3}{a - 2} = \frac{3 - 3(a - 1)}{(a - 1)(a - 2)} = \frac{3 - 3a + 3}{(a - 1)(a - 2)} = \frac{6 - 3a}{(a - 1)(a - 2)} = \frac{3(2 - a)}{(a - 1)(a - 2)} = -\frac{3(a - 2)}{(a - 1)(a - 2)} = -\frac{3}{a - 1} = \frac{3}{1 - a}$

Решение 2

$\frac{b^2 + 3b}{b^3 + 9b} * (\frac{b - 3}{b + 3} + \frac{b + 3}{b - 3}) = \frac{b(b + 3)}{b(b^2 + 9)} * \frac{(b - 3)^2 + (b + 3)^2}{(b - 3)(b + 3)} = \frac{b + 3}{b^2 + 9} * \frac{b^2 - 6b + 9 + b^2 + 6b + 9}{(b - 3)(b + 3)} = \frac{1}{b^2 + 9} * \frac{2b^2 + 18}{b - 3} = \frac{1}{b^2 + 9} * \frac{2(b^2 + 9)}{b - 3} = \frac{1}{1} * \frac{2}{b - 3} = \frac{2}{b - 3}$

Решение 3

$(\frac{3c + 1}{3c - 1} - \frac{3c - 1}{3c + 1}) : \frac{2c}{6c + 2} = \frac{(3c + 1)^2 - (3c - 1)^2}{(3c - 1)(3c + 1)} : \frac{2c}{2(3c + 1)} = \frac{9c^2 + 6c + 1 - (9c^2 - 6c + 1)}{(3c - 1)(3c + 1)} * \frac{2(3c + 1)}{2c} = \frac{12c}{3c - 1} * \frac{2}{2c} = \frac{6}{3c - 1} * \frac{2}{1} = \frac{12}{3c - 1}$

Решение 4

$(\frac{1}{a^2 - 4ab + 4b^2} - \frac{1}{4b^2 - a^2}) : \frac{2a}{a^2 - 4b^2} = (\frac{1}{(a - 2b)^2} - \frac{1}{(2b - a)(2b + a)}) : \frac{2a}{(a - 2b)(a + 2b)} = (\frac{1}{(2b - a)^2} - \frac{1}{(2b - a)(2b + a)}) : \frac{2a}{(a - 2b)(a + 2b)} = \frac{2b + a - (2b - a)}{(2b - a)^2(2b + a)} * \frac{(a - 2b)(a + 2b)}{2a} = \frac{2b + a - 2b + a}{(a - 2b)^2(a + 2b)} * \frac{(a - 2b)(a + 2b)}{2a} = \frac{2a}{(a - 2b)^2(a + 2b)} * \frac{(a - 2b)(a + 2b)}{2a} = \frac{1}{a - 2b} * \frac{1}{1} = \frac{1}{a - 2b}$

Решение 5

$(\frac{a - 8}{a^2 - 10a + 25} - \frac{a}{a^2 - 25}) : \frac{a - 20}{(a - 5)^2} = (\frac{a - 8}{(a - 5)^2} - \frac{a}{(a - 5)(a + 5)}) : \frac{a - 20}{(a - 5)^2} = \frac{(a - 8)(a + 5) - a(a - 5)}{(a - 5)^2(a + 5)} * \frac{(a - 5)^2}{a - 20} = \frac{a^2 - 8a + 5a - 40 - a^2 + 5a}{a + 5} * \frac{1}{a - 20} = \frac{2a - 40}{a + 5} * \frac{1}{a - 20} = \frac{2(a - 20)}{a + 5} * \frac{1}{a - 20} = \frac{2}{a + 5}$

Решение 6

$(\frac{2x + 1}{x^2 + 6x + 9} - \frac{x - 2}{x^2 + 3x}) : \frac{x^2 + 6}{x^3 - 9x} = (\frac{2x + 1}{(x + 3)^2} - \frac{x - 2}{x(x + 3)}) : \frac{x^2 + 6}{x(x^2 - 9)} = \frac{x(2x + 1) - (x - 2)(x + 3)}{x(x + 3)^2} : \frac{x^2 + 6}{x(x - 3)(x + 3)} = \frac{2x^2 + x - (x^2 - 2x + 3x - 6)}{x(x + 3)^2} * \frac{x(x - 3)(x + 3)}{x^2 + 6} = \frac{2x^2 + x - x^2 + 2x - 3x + 6}{x(x + 3)} * \frac{x(x - 3)}{x^2 + 6} = \frac{x^2 + 6}{x(x + 3)} * \frac{x(x - 3)}{x^2 + 6} = \frac{1}{x + 3} * \frac{x - 3}{1} = \frac{x - 3}{x + 3}$