Предмет: Математика
1. Выполните действия:
а) $$
\frac{15a^6}{b^7} : \frac{2a^3}{b^6} = \frac{15a^6}{b^7} \cdot \frac{b^6}{2a^3} = \frac{15a^3}{2b}
$$
б) $$
\frac{21m^5}{25-10m+m^2} : \frac{5-m}{7m^2} = \frac{21m^5}{(5-m)^2} \cdot \frac{7m^2}{5-m} = \frac{147m^7}{(5-m)^3}
$$
в) $$
\frac{3x+6}{x+3} \cdot \frac{x^2-9}{x^2-4} = \frac{3(x+2)}{x+3} \cdot \frac{(x-3)(x+3)}{(x-2)(x+2)} = \frac{3(x-3)}{x-2}
$$
2. Упростите выражение:
$$
\frac{4x^2+2x+6}{x^2-1} : (\frac{3}{x-1} + \frac{2x}{x+1}) = \frac{4x^2+2x+6}{(x-1)(x+1)} : \frac{3(x+1) + 2x(x-1)}{(x-1)(x+1)} = \frac{4x^2+2x+6}{(x-1)(x+1)} : \frac{3x+3 + 2x^2-2x}{(x-1)(x+1)} = \frac{4x^2+2x+6}{(x-1)(x+1)} : \frac{2x^2+x+3}{(x-1)(x+1)} = \frac{4x^2+2x+6}{(x-1)(x+1)} \cdot \frac{(x-1)(x+1)}{2x^2+x+3} = \frac{2(2x^2+x+3)}{2x^2+x+3} = 2
$$
3. Докажите тождество:
$$
\frac{b}{b-a} + \frac{a^2-b^2}{ab+a^2} - \frac{b^2}{(b-a)^2} = -\frac{b}{a}
$$
$$
\frac{b}{b-a} + \frac{(a-b)(a+b)}{a(b+a)} - \frac{b^2}{(b-a)^2} = \frac{b}{b-a} + \frac{a-b}{a} - \frac{b^2}{(b-a)^2} = \frac{ab(b-a) + (a-b)(b-a)^2 - ab^2}{a(b-a)^2} = \frac{ab^2 - a^2b + (a-b)(b^2 - 2ab + a^2) - ab^2}{a(b-a)^2} = \frac{-a^2b + ab^2 - 2a^2b + a^3 - b^3 + 2ab^2 - a^2b - ab^2}{a(b-a)^2} = \frac{-4a^2b + a^3 + 2ab^2 - b^3}{a(b-a)^2} = \frac{a^3-b^3-3ba^2+3a^2b}{a(b-a)^2} = \frac{-a(b-a)^2}{a(b-a)^2} = -\frac{b}{a}
$$
4. Найдите значение \(x\), при котором значение дроби \(\frac{12}{x}\) меньше значения дроби \(\frac{x}{x-8}\) на 1.
$$
\frac{x}{x-8} - \frac{12}{x} = 1
$$
$$
\frac{x^2 - 12(x-8)}{x(x-8)} = 1
$$
$$
x^2 - 12x + 96 = x^2 - 8x
$$
$$
-12x + 96 = -8x
$$
$$
4x = 96
$$
$$
x = 24
$$