Страницы 45-46 номер 2.3.27, ГДЗ по алгебре за 7, 8 и 9 класс к задачнику Ткачевой
Выполнить действия: 1) \(\left(\frac{a}{a^2-6a+9}-\frac{a+2}{a^2-a-6}\right)\cdot(2a-6)^2\); 2) \(\left(\frac{1}{b^2-1}-\frac{b-1}{2b^2+3b+1}\right):\frac{b-4}{b^2-1}\); 3) \(\frac{9x^2-4}{6x^2-x-2}:\frac{3x+2}{2x^2-5x-3}+\frac{4x-1}{3-x}\); 4) \(\frac{2y}{y-6}+\frac{25y^2-16}{10y^2+3y-4}\cdot\frac{1-2y}{5y^2-4y}\).
1) \[\begin{aligned} &\frac{a}{a^2 - 6a + 9} - \frac{a + 2}{a^2 - a - 6} = \frac{a}{(a - 3)^2} - \frac{a + 2}{(a - 3)(a + 2)} = {} \\ &= \frac{a}{(a - 3)^2} - \frac{1}{a - 3} = \frac{a - (a - 3)}{(a - 3)^2} = \frac{3}{(a - 3)^2}; \end{aligned}\]
\[\frac{3}{(a - 3)^2} \cdot (2a - 6)^2 = \frac{3}{(a - 3)^2} \cdot 4(a - 3)^2 = 12.\]
2) \(b^2 - 1 = (b - 1)(b + 1)\), \(2b^2 + 3b + 1 = 2b^2 + 2b + b + 1 = 2b(b + 1) + (b + 1) = (b + 1)(2b + 1)\), поэтому
\[\begin{aligned} &\frac{1}{b^2 - 1} - \frac{b - 1}{2b^2 + 3b + 1} = \frac{1}{(b - 1)(b + 1)} - \frac{b - 1}{(b + 1)(2b + 1)} = {} \\ &= \frac{2b + 1 - (b - 1)^2}{(b - 1)(b + 1)(2b + 1)} = \frac{2b + 1 - b^2 + 2b - 1}{(b - 1)(b + 1)(2b + 1)} = {} \\ &= \frac{4b - b^2}{(b - 1)(b + 1)(2b + 1)} = \frac{-b(b - 4)}{(b - 1)(b + 1)(2b + 1)}; \end{aligned}\]
\[\frac{-b(b - 4)}{(b - 1)(b + 1)(2b + 1)} : \frac{b - 4}{b^2 - 1} = \frac{-b(b - 4)}{(b - 1)(b + 1)(2b + 1)} \cdot \frac{(b - 1)(b + 1)}{b - 4} = -\frac{b}{2b + 1}.\]
3) \(9x^2 - 4 = (3x - 2)(3x + 2)\), \(6x^2 - x - 2 = (3x - 2)(2x + 1)\), \(2x^2 - 5x - 3 = (x - 3)(2x + 1)\), поэтому
\[\frac{9x^2 - 4}{6x^2 - x - 2} : \frac{3x + 2}{2x^2 - 5x - 3} = \frac{(3x - 2)(3x + 2)}{(3x - 2)(2x + 1)} \cdot \frac{(x - 3)(2x + 1)}{3x + 2} = x - 3.\]
В дроби \(\frac{4x - 1}{3 - x}\) меняем знак знаменателя и знак перед дробью:
\[\begin{aligned} &x - 3 + \frac{4x - 1}{3 - x} = x - 3 - \frac{4x - 1}{x - 3} = \frac{(x - 3)^2 - (4x - 1)}{x - 3} = {} \\ &= \frac{x^2 - 6x + 9 - 4x + 1}{x - 3} = \frac{x^2 - 10x + 10}{x - 3}. \end{aligned}\]
4) \(25y^2 - 16 = (5y - 4)(5y + 4)\), \(10y^2 + 3y - 4 = (5y + 4)(2y - 1)\), \(5y^2 - 4y = y(5y - 4)\), поэтому
\[\begin{aligned} &\frac{25y^2 - 16}{10y^2 + 3y - 4} \cdot \frac{1 - 2y}{5y^2 - 4y} = \frac{(5y - 4)(5y + 4)(1 - 2y)}{(5y + 4)(2y - 1) \cdot y(5y - 4)} = {} \\ &= \frac{1 - 2y}{y(2y - 1)} = \frac{-(2y - 1)}{y(2y - 1)} = -\frac{1}{y}; \end{aligned}\]
\[\frac{2y}{y - 6} - \frac{1}{y} = \frac{2y \cdot y - (y - 6)}{y(y - 6)} = \frac{2y^2 - y + 6}{y(y - 6)}.\]
Ответ: 1) \(12\); 2) \(-\frac{b}{2b + 1}\); 3) \(\frac{x^2 - 10x + 10}{x - 3}\); 4) \(\frac{2y^2 - y + 6}{y(y - 6)}\).
Выполнить действия: 1) \(\left(\frac{a}{a^2-6a+9}-\frac{a+2}{a^2-a-6}\right)\cdot(2a-6)^2\); 2) \(\left(\frac{1}{b^2-1}-\frac{b-1}{2b^2+3b+1}\right):\frac{b-4}{b^2-1}\); 3) \(\frac{9x^2-4}{6x^2-x-2}:\frac{3x+2}{2x^2-5x-3}+\frac{4x-1}{3-x}\); 4) \(\frac{2y}{y-6}+\frac{25y^2-16}{10y^2+3y-4}\cdot\frac{1-2y}{5y^2-4y}\).
1) \(a^2 - 6a + 9 = (a - 3)^2\), \(a^2 - a - 6 = (a - 3)(a + 2)\), \((2a - 6)^2 = 4(a - 3)^2\), поэтому
\[\begin{aligned} &\frac{a}{a^2 - 6a + 9} - \frac{a + 2}{a^2 - a - 6} = \frac{a}{(a - 3)^2} - \frac{a + 2}{(a - 3)(a + 2)} = {} \\ &= \frac{a}{(a - 3)^2} - \frac{1}{a - 3} = \frac{a - (a - 3)}{(a - 3)^2} = \frac{3}{(a - 3)^2}; \end{aligned}\]
\[\frac{3}{(a - 3)^2} \cdot (2a - 6)^2 = \frac{3}{(a - 3)^2} \cdot 4(a - 3)^2 = 12.\]
2) \(b^2 - 1 = (b - 1)(b + 1)\), \(2b^2 + 3b + 1 = 2b^2 + 2b + b + 1 = 2b(b + 1) + (b + 1) = (b + 1)(2b + 1)\), поэтому
\[\begin{aligned} &\frac{1}{b^2 - 1} - \frac{b - 1}{2b^2 + 3b + 1} = \frac{1}{(b - 1)(b + 1)} - \frac{b - 1}{(b + 1)(2b + 1)} = {} \\ &= \frac{2b + 1 - (b - 1)^2}{(b - 1)(b + 1)(2b + 1)} = \frac{2b + 1 - b^2 + 2b - 1}{(b - 1)(b + 1)(2b + 1)} = {} \\ &= \frac{4b - b^2}{(b - 1)(b + 1)(2b + 1)} = \frac{-b(b - 4)}{(b - 1)(b + 1)(2b + 1)}; \end{aligned}\]
\[\frac{-b(b - 4)}{(b - 1)(b + 1)(2b + 1)} : \frac{b - 4}{b^2 - 1} = \frac{-b(b - 4)}{(b - 1)(b + 1)(2b + 1)} \cdot \frac{(b - 1)(b + 1)}{b - 4} = -\frac{b}{2b + 1}.\]
3) \(9x^2 - 4 = (3x - 2)(3x + 2)\), \(6x^2 - x - 2 = (3x - 2)(2x + 1)\), \(2x^2 - 5x - 3 = (x - 3)(2x + 1)\), поэтому
\[\frac{9x^2 - 4}{6x^2 - x - 2} : \frac{3x + 2}{2x^2 - 5x - 3} = \frac{(3x - 2)(3x + 2)}{(3x - 2)(2x + 1)} \cdot \frac{(x - 3)(2x + 1)}{3x + 2} = x - 3.\]
В дроби \(\frac{4x - 1}{3 - x}\) меняем знак знаменателя и знак перед дробью:
\[\begin{aligned} &x - 3 + \frac{4x - 1}{3 - x} = x - 3 - \frac{4x - 1}{x - 3} = \frac{(x - 3)^2 - (4x - 1)}{x - 3} = {} \\ &= \frac{x^2 - 6x + 9 - 4x + 1}{x - 3} = \frac{x^2 - 10x + 10}{x - 3}. \end{aligned}\]
4) \(25y^2 - 16 = (5y - 4)(5y + 4)\), \(10y^2 + 3y - 4 = (5y + 4)(2y - 1)\), \(5y^2 - 4y = y(5y - 4)\), поэтому
\[\begin{aligned} &\frac{25y^2 - 16}{10y^2 + 3y - 4} \cdot \frac{1 - 2y}{5y^2 - 4y} = \frac{(5y - 4)(5y + 4)(1 - 2y)}{(5y + 4)(2y - 1) \cdot y(5y - 4)} = {} \\ &= \frac{1 - 2y}{y(2y - 1)} = \frac{-(2y - 1)}{y(2y - 1)} = -\frac{1}{y}; \end{aligned}\]
\[\frac{2y}{y - 6} - \frac{1}{y} = \frac{2y \cdot y - (y - 6)}{y(y - 6)} = \frac{2y^2 - y + 6}{y(y - 6)}.\]
Ответ: 1) \(12\); 2) \(-\frac{b}{2b + 1}\); 3) \(\frac{x^2 - 10x + 10}{x - 3}\); 4) \(\frac{2y^2 - y + 6}{y(y - 6)}\).
