Страница 44 номер 2.3.23, ГДЗ по алгебре за 7, 8 и 9 класс к задачнику Ткачевой
Упростить выражение: 1) \(\frac{a^2+2ab+b^2}{a^2-b^2}\cdot(3a+3b):(a^2b+ab^2)\); 2) \(\frac{25x^2-y^2}{25x^2-10xy+y^2}:(5x^2+xy)\cdot(y-5x)\); 3) \(\frac{x^3+y^3}{15xy^2}\cdot\frac{20x^2y}{x^2-xy+y^2}:\frac{x+y}{9x^3y^3}\); 4) \(\frac{16a^2+12a+9}{24a^3b^5}:\frac{27-64a^3}{16a^4b}\cdot\frac{4a-3}{2ab^3}\).
1) \[\begin{aligned} &\frac{a^2 + 2ab + b^2}{a^2 - b^2} \cdot (3a + 3b) : (a^2b + ab^2) = {} \\ &= \frac{(a + b)^2}{(a - b)(a + b)} \cdot \frac{3(a + b)}{1} \cdot \frac{1}{ab(a + b)} = {} \\ &= \frac{(a + b)^2 \cdot 3(a + b)}{(a - b)(a + b) \cdot ab(a + b)} = \frac{3(a + b)}{ab(a - b)}; \end{aligned}\]
2) \[\begin{aligned} &\frac{25x^2 - y^2}{25x^2 - 10xy + y^2} : (5x^2 + xy) \cdot (y - 5x) = {} \\ &= \frac{(5x - y)(5x + y)}{(5x - y)^2} \cdot \frac{1}{x(5x + y)} \cdot \frac{y - 5x}{1} = {} \\ &= \frac{(5x - y)(5x + y)(y - 5x)}{(5x - y)^2 \cdot x(5x + y)} = \frac{y - 5x}{x(5x - y)} = {} \\ &= \frac{-(5x - y)}{x(5x - y)} = -\frac{1}{x}; \end{aligned}\]
3) \[\begin{aligned} &\frac{x^3 + y^3}{15xy^2} \cdot \frac{20x^2y}{x^2 - xy + y^2} : \frac{x + y}{9x^3y^3} = {} \\ &= \frac{(x + y)(x^2 - xy + y^2)}{15xy^2} \cdot \frac{20x^2y}{x^2 - xy + y^2} \cdot \frac{9x^3y^3}{x + y} = {} \\ &= \frac{(x + y)(x^2 - xy + y^2) \cdot 20x^2y \cdot 9x^3y^3}{15xy^2 \cdot (x^2 - xy + y^2)(x + y)} = 12x^4y^2; \end{aligned}\]
4) \[\begin{aligned} &\frac{16a^2 + 12a + 9}{24a^3b^5} : \frac{27 - 64a^3}{16a^4b} \cdot \frac{4a - 3}{2ab^3} = {} \\ &= \frac{16a^2 + 12a + 9}{24a^3b^5} \cdot \frac{16a^4b}{(3 - 4a)(9 + 12a + 16a^2)} \cdot \frac{4a - 3}{2ab^3} = {} \\ &= \frac{16a^4b(4a - 3)}{24a^3b^5 (3 - 4a) \cdot 2ab^3} = -\frac{16a^4b}{48a^4b^8} = -\frac{1}{3b^7}. \end{aligned}\]
Ответ: 1) \(\frac{3(a + b)}{ab(a - b)}\); 2) \(-\frac{1}{x}\); 3) \(12x^4y^2\); 4) \(-\frac{1}{3b^7}\).
Упростить выражение: 1) \(\frac{a^2+2ab+b^2}{a^2-b^2}\cdot(3a+3b):(a^2b+ab^2)\); 2) \(\frac{25x^2-y^2}{25x^2-10xy+y^2}:(5x^2+xy)\cdot(y-5x)\); 3) \(\frac{x^3+y^3}{15xy^2}\cdot\frac{20x^2y}{x^2-xy+y^2}:\frac{x+y}{9x^3y^3}\); 4) \(\frac{16a^2+12a+9}{24a^3b^5}:\frac{27-64a^3}{16a^4b}\cdot\frac{4a-3}{2ab^3}\).
Деление заменяется умножением на дробь, обратную делителю, а многочлен записывается дробью со знаменателем 1.
1) \[\begin{aligned} &\frac{a^2 + 2ab + b^2}{a^2 - b^2} \cdot (3a + 3b) : (a^2b + ab^2) = {} \\ &= \frac{(a + b)^2}{(a - b)(a + b)} \cdot \frac{3(a + b)}{1} \cdot \frac{1}{ab(a + b)} = {} \\ &= \frac{(a + b)^2 \cdot 3(a + b)}{(a - b)(a + b) \cdot ab(a + b)} = \frac{3(a + b)}{ab(a - b)}; \end{aligned}\]
2) \[\begin{aligned} &\frac{25x^2 - y^2}{25x^2 - 10xy + y^2} : (5x^2 + xy) \cdot (y - 5x) = {} \\ &= \frac{(5x - y)(5x + y)}{(5x - y)^2} \cdot \frac{1}{x(5x + y)} \cdot \frac{y - 5x}{1} = {} \\ &= \frac{(5x - y)(5x + y)(y - 5x)}{(5x - y)^2 \cdot x(5x + y)} = \frac{y - 5x}{x(5x - y)} = {} \\ &= \frac{-(5x - y)}{x(5x - y)} = -\frac{1}{x}; \end{aligned}\]
3) \[\begin{aligned} &\frac{x^3 + y^3}{15xy^2} \cdot \frac{20x^2y}{x^2 - xy + y^2} : \frac{x + y}{9x^3y^3} = {} \\ &= \frac{(x + y)(x^2 - xy + y^2)}{15xy^2} \cdot \frac{20x^2y}{x^2 - xy + y^2} \cdot \frac{9x^3y^3}{x + y} = {} \\ &= \frac{(x + y)(x^2 - xy + y^2) \cdot 20x^2y \cdot 9x^3y^3}{15xy^2 \cdot (x^2 - xy + y^2)(x + y)} = 12x^4y^2; \end{aligned}\]
4) \[\begin{aligned} &\frac{16a^2 + 12a + 9}{24a^3b^5} : \frac{27 - 64a^3}{16a^4b} \cdot \frac{4a - 3}{2ab^3} = {} \\ &= \frac{16a^2 + 12a + 9}{24a^3b^5} \cdot \frac{16a^4b}{(3 - 4a)(9 + 12a + 16a^2)} \cdot \frac{4a - 3}{2ab^3} = {} \\ &= \frac{16a^4b(4a - 3)}{24a^3b^5 (3 - 4a) \cdot 2ab^3} = -\frac{16a^4b}{48a^4b^8} = -\frac{1}{3b^7}. \end{aligned}\]
Ответ: 1) \(\frac{3(a + b)}{ab(a - b)}\); 2) \(-\frac{1}{x}\); 3) \(12x^4y^2\); 4) \(-\frac{1}{3b^7}\).
