Страница 45 номер 2.3.25, ГДЗ по алгебре за 7, 8 и 9 класс к задачнику Ткачевой
Сократить дробь: 1) \(\frac{2x^2+5x+2}{2x+1}\); 2) \(\frac{x+1}{3x^2+5x+2}\); 3) \(\frac{3x^2+5x-2}{3x+6}\); 4) \(\frac{10x-5}{2x^2+3x-2}\); 5) \(\frac{x^2+2x+1}{4x^2+5x+1}\); 6) \(\frac{3x^2-4x+1}{x^2-2x+1}\); 7) \(\frac{7x-2x^2-3}{2x^2-x}\); 8) \(\frac{2+x-3x^2}{9x^2-4}\).
1) \[\begin{aligned} &\frac{2x^2 + 5x + 2}{2x + 1} = \frac{2x^2 + 4x + x + 2}{2x + 1} = {} \\ &= \frac{2x(x + 2) + (x + 2)}{2x + 1} = \frac{(x + 2)(2x + 1)}{2x + 1} = x + 2; \end{aligned}\]
2) \[\begin{aligned} &\frac{x + 1}{3x^2 + 5x + 2} = \frac{x + 1}{3x^2 + 3x + 2x + 2} = {} \\ &= \frac{x + 1}{3x(x + 1) + 2(x + 1)} = \frac{x + 1}{(x + 1)(3x + 2)} = \frac{1}{3x + 2}; \end{aligned}\]
3) \[\begin{aligned} &\frac{3x^2 + 5x - 2}{3x + 6} = \frac{3x^2 + 6x - x - 2}{3(x + 2)} = {} \\ &= \frac{3x(x + 2) - (x + 2)}{3(x + 2)} = \frac{(x + 2)(3x - 1)}{3(x + 2)} = \frac{3x - 1}{3}; \end{aligned}\]
4) \[\begin{aligned} &\frac{10x - 5}{2x^2 + 3x - 2} = \frac{5(2x - 1)}{2x^2 + 4x - x - 2} = {} \\ &= \frac{5(2x - 1)}{2x(x + 2) - (x + 2)} = \frac{5(2x - 1)}{(x + 2)(2x - 1)} = \frac{5}{x + 2}; \end{aligned}\]
5) \[\begin{aligned} &\frac{x^2 + 2x + 1}{4x^2 + 5x + 1} = \frac{(x + 1)^2}{4x^2 + 4x + x + 1} = {} \\ &= \frac{(x + 1)^2}{4x(x + 1) + (x + 1)} = \frac{(x + 1)^2}{(x + 1)(4x + 1)} = \frac{x + 1}{4x + 1}; \end{aligned}\]
6) \[\begin{aligned} &\frac{3x^2 - 4x + 1}{x^2 - 2x + 1} = \frac{3x^2 - 3x - x + 1}{(x - 1)^2} = {} \\ &= \frac{3x(x - 1) - (x - 1)}{(x - 1)^2} = \frac{(x - 1)(3x - 1)}{(x - 1)^2} = \frac{3x - 1}{x - 1}; \end{aligned}\]
7) \[\begin{aligned} &\frac{7x - 2x^2 - 3}{2x^2 - x} = \frac{-(2x^2 - 7x + 3)}{x(2x - 1)} = \frac{-(2x^2 - 6x - x + 3)}{x(2x - 1)} = {} \\ &= \frac{-(2x(x - 3) - (x - 3))}{x(2x - 1)} = \frac{-(x - 3)(2x - 1)}{x(2x - 1)} = \frac{3 - x}{x}; \end{aligned}\]
8) \[\begin{aligned} &\frac{2 + x - 3x^2}{9x^2 - 4} = \frac{-(3x^2 - x - 2)}{(3x - 2)(3x + 2)} = \frac{-(3x^2 - 3x + 2x - 2)}{(3x - 2)(3x + 2)} = {} \\ &= \frac{-(3x(x - 1) + 2(x - 1))}{(3x - 2)(3x + 2)} = \frac{-(x - 1)(3x + 2)}{(3x - 2)(3x + 2)} = \frac{1 - x}{3x - 2}. \end{aligned}\]
Ответ: 1) \(x + 2\); 2) \(\frac{1}{3x + 2}\); 3) \(\frac{3x - 1}{3}\); 4) \(\frac{5}{x + 2}\); 5) \(\frac{x + 1}{4x + 1}\); 6) \(\frac{3x - 1}{x - 1}\); 7) \(\frac{3 - x}{x}\); 8) \(\frac{1 - x}{3x - 2}\).
Сократить дробь: 1) \(\frac{2x^2+5x+2}{2x+1}\); 2) \(\frac{x+1}{3x^2+5x+2}\); 3) \(\frac{3x^2+5x-2}{3x+6}\); 4) \(\frac{10x-5}{2x^2+3x-2}\); 5) \(\frac{x^2+2x+1}{4x^2+5x+1}\); 6) \(\frac{3x^2-4x+1}{x^2-2x+1}\); 7) \(\frac{7x-2x^2-3}{2x^2-x}\); 8) \(\frac{2+x-3x^2}{9x^2-4}\).
Дробь сокращают только после разложения её числителя и знаменателя на множители. Квадратный трёхчлен раскладывают так: его средний член представляют в виде суммы двух одночленов и применяют способ группировки.
1) \[\begin{aligned} &\frac{2x^2 + 5x + 2}{2x + 1} = \frac{2x^2 + 4x + x + 2}{2x + 1} = {} \\ &= \frac{2x(x + 2) + (x + 2)}{2x + 1} = \frac{(x + 2)(2x + 1)}{2x + 1} = x + 2; \end{aligned}\]
2) \[\begin{aligned} &\frac{x + 1}{3x^2 + 5x + 2} = \frac{x + 1}{3x^2 + 3x + 2x + 2} = {} \\ &= \frac{x + 1}{3x(x + 1) + 2(x + 1)} = \frac{x + 1}{(x + 1)(3x + 2)} = \frac{1}{3x + 2}; \end{aligned}\]
3) \[\begin{aligned} &\frac{3x^2 + 5x - 2}{3x + 6} = \frac{3x^2 + 6x - x - 2}{3(x + 2)} = {} \\ &= \frac{3x(x + 2) - (x + 2)}{3(x + 2)} = \frac{(x + 2)(3x - 1)}{3(x + 2)} = \frac{3x - 1}{3}; \end{aligned}\]
4) \[\begin{aligned} &\frac{10x - 5}{2x^2 + 3x - 2} = \frac{5(2x - 1)}{2x^2 + 4x - x - 2} = {} \\ &= \frac{5(2x - 1)}{2x(x + 2) - (x + 2)} = \frac{5(2x - 1)}{(x + 2)(2x - 1)} = \frac{5}{x + 2}; \end{aligned}\]
5) \[\begin{aligned} &\frac{x^2 + 2x + 1}{4x^2 + 5x + 1} = \frac{(x + 1)^2}{4x^2 + 4x + x + 1} = {} \\ &= \frac{(x + 1)^2}{4x(x + 1) + (x + 1)} = \frac{(x + 1)^2}{(x + 1)(4x + 1)} = \frac{x + 1}{4x + 1}; \end{aligned}\]
6) \[\begin{aligned} &\frac{3x^2 - 4x + 1}{x^2 - 2x + 1} = \frac{3x^2 - 3x - x + 1}{(x - 1)^2} = {} \\ &= \frac{3x(x - 1) - (x - 1)}{(x - 1)^2} = \frac{(x - 1)(3x - 1)}{(x - 1)^2} = \frac{3x - 1}{x - 1}; \end{aligned}\]
7) \[\begin{aligned} &\frac{7x - 2x^2 - 3}{2x^2 - x} = \frac{-(2x^2 - 7x + 3)}{x(2x - 1)} = \frac{-(2x^2 - 6x - x + 3)}{x(2x - 1)} = {} \\ &= \frac{-(2x(x - 3) - (x - 3))}{x(2x - 1)} = \frac{-(x - 3)(2x - 1)}{x(2x - 1)} = \frac{3 - x}{x}; \end{aligned}\]
8) \[\begin{aligned} &\frac{2 + x - 3x^2}{9x^2 - 4} = \frac{-(3x^2 - x - 2)}{(3x - 2)(3x + 2)} = \frac{-(3x^2 - 3x + 2x - 2)}{(3x - 2)(3x + 2)} = {} \\ &= \frac{-(3x(x - 1) + 2(x - 1))}{(3x - 2)(3x + 2)} = \frac{-(x - 1)(3x + 2)}{(3x - 2)(3x + 2)} = \frac{1 - x}{3x - 2}. \end{aligned}\]
Каждое из полученных равенств верно при всех значениях \(x\), при которых обе его части имеют смысл.
Ответ: 1) \(x + 2\); 2) \(\frac{1}{3x + 2}\); 3) \(\frac{3x - 1}{3}\); 4) \(\frac{5}{x + 2}\); 5) \(\frac{x + 1}{4x + 1}\); 6) \(\frac{3x - 1}{x - 1}\); 7) \(\frac{3 - x}{x}\); 8) \(\frac{1 - x}{3x - 2}\).
