Homepage › Solution manuals › Найма Гахраманова › Математика - 8 › Задание 9, стр 114

Задание 9, стр 114

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\begin{aligned} a) & x^{2} - x - 6 = 0 \\ D = b^{2} - 4 ac = 1 - 4 \cdot (- 6) = 1 + 24 = 25 = 5^{2} . \\ x_{1} = \frac{1 + 5}{2} = \frac{6}{2} = 3; x_{2} = \frac{- b - \sqrt{2}}{2 a} = \frac{1 - 5}{2} = \frac{- 4}{2} = - 2 . \\ \Rightarrow x^{2} - x - 6 = (x - 3) (x + 2) . \\ \frac{x^{2} - x - 6}{4 x + 4} \cdot \frac{x + 1}{x^{2} - 6 x + 9} = \frac{(x - 3) (x + 2)}{4 (x + 1)} \cdot \frac{x + 1}{{(x - 3)}^{2}} = \\ = \frac{x + 2}{4 (x - 3)} = \frac{x + 2}{4 x - 12} . \\ b) & \frac{2 d + 8}{d^{2} + 5 d + 6} : \frac{d + 4}{d^{2} - 4} = \frac{2 (d + 4)}{d^{2} + 5 d + 6} \cdot \frac{d^{2} - 4}{d + 4} = \frac{2 (d + 4)}{d^{2} + 5 d + 6} \cdot \frac{(d - 2) (d + 2)}{d + 4} = \frac{2 (d - 2) (d + 2)}{d^{2} + 5 d + 6} . \\ d^{2} + 5 d + 6 = 0 \\ D = b^{2} - 4 ac = 5^{2} - 4 \cdot 6 = 25 - 24 = 1^{2} . \\ d_{1} = \frac{- b + \sqrt{2}}{2 a} = \frac{- 5 + 1}{2} = \frac{- 4}{2} = - 2; d 2 = \frac{- 5 - 1}{2} = \frac{- 6}{2} = - 3 \\ \Rightarrow d^{2} + 5 d + 6 = (d + 2) (d + 3) \\ \frac{2 (d - 2) (d + 2)}{d d^{2} + 5 d + 6} = \frac{2 (d - 2) (d + 2)}{(d + 2) (d + 3)} = \frac{2 (d - 2)}{d + 3} \\ c) & \frac{x^{3} + 5 x^{2} - x - 5}{x^{2} - 25} : (x + 1) = \frac{x^{2} (x + 5) - (x + 5)}{(x - 5) (x + 5)} \cdot \frac{1}{x + 1} = \frac{(x^{2} - 1) (x + 5)}{(x - 5) (x + 5)} \cdot \frac{1}{x + 1} = \frac{(x - 1) (x + 1) (x + 5)}{(x - 5) (x + 5)} \cdot \frac{1}{x + 1} = \frac{x - 1}{x - 5} \\ d) & \frac{n^{3} - m^{3}}{n^{2} - m^{2}} \cdot \frac{n^{2} + 2 mn + m^{2}}{n^{2} + nm + m^{2}} = \frac{(n - m) (n^{2} + nm + m^{2})}{(n - m) (n + m)} \cdot \frac{{(m + n)}^{2}}{n^{2} + nm + m^{2}} = m + n . \end{aligned}

gimli

2023-01-17 12:47

Задание 9, стр 114

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