Ответ:
на фото
Объяснение:
По теореме Виета:
[tex] {x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c [/tex]
[tex] {a}^{2} + 5a - 14 = 0 \\ a_{1} + a_{2} = - 5 \\ a_{1}a_{2} = - 14 \\ a_{1} = - 7 \\ a_{2} = 2 \\ a {}^{2} - 5a - 14 = (a + 7)(a - 2) \\ \\ a {}^{2} + 14a + 49 = (a + 7) {}^{2} \\ \\ \frac{ {a}^{2} + 5a - 14 }{a {}^{2} + 14a + 49 } = \frac{(a + 7)(a - 2)}{(a + 7) {}^{2} } = \frac{a - 2}{a + 7} [/tex]
[tex] {x}^{2} - 11x + 24 = 0 \\ x_{1} + x_{2} = 11\\ x_{1} x_{2} = 24 \\ x_{1} =8 \\ x_{2} = 3 \\ {x}^{2} - 11x + 24 = (x - 8)(x - 3) \\ \\ {x}^{2} - x - 6 = 0 \\ x_{1} + x_{2} =1 \\ x_{1} x_{2} = - 6\\ x_{1} = 3\\ x_{2} = - 2 \\ {x}^{2} - x - 6 = (x - 3)(x +2 ) \\ \\ \frac{ {x}^{2} - 11x + 24 }{ {x}^{2} - x - 6} = \frac{(x - 8)(x - 3)}{(x - 3)(x + 2)} = \frac{x - 8}{x + 2} [/tex]
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Answers & Comments
Ответ:
на фото
Объяснение:
По теореме Виета:
[tex] {x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c [/tex]
а)
[tex] {a}^{2} + 5a - 14 = 0 \\ a_{1} + a_{2} = - 5 \\ a_{1}a_{2} = - 14 \\ a_{1} = - 7 \\ a_{2} = 2 \\ a {}^{2} - 5a - 14 = (a + 7)(a - 2) \\ \\ a {}^{2} + 14a + 49 = (a + 7) {}^{2} \\ \\ \frac{ {a}^{2} + 5a - 14 }{a {}^{2} + 14a + 49 } = \frac{(a + 7)(a - 2)}{(a + 7) {}^{2} } = \frac{a - 2}{a + 7} [/tex]
б)
[tex] {x}^{2} - 11x + 24 = 0 \\ x_{1} + x_{2} = 11\\ x_{1} x_{2} = 24 \\ x_{1} =8 \\ x_{2} = 3 \\ {x}^{2} - 11x + 24 = (x - 8)(x - 3) \\ \\ {x}^{2} - x - 6 = 0 \\ x_{1} + x_{2} =1 \\ x_{1} x_{2} = - 6\\ x_{1} = 3\\ x_{2} = - 2 \\ {x}^{2} - x - 6 = (x - 3)(x +2 ) \\ \\ \frac{ {x}^{2} - 11x + 24 }{ {x}^{2} - x - 6} = \frac{(x - 8)(x - 3)}{(x - 3)(x + 2)} = \frac{x - 8}{x + 2} [/tex]