Question

Решите пж пж пж срочно надо
8x^2-x-9=0
x^4-16x^2+28=0 ^-степень 2 решить биквадратное
пж пж пж пж пж

Answer 1

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$8 {x}^{2} - x - 9 = 0 \\ d = 1 - 4 \times 8 \times ( - 9) = 1 + 288 = 289 \\ x1 = \frac{1 + \sqrt{d} }{2 \times 8} = \frac{1 + \sqrt{289} }{16} = \frac{1 + 17}{16} = \frac{18}{16} = \frac{9}{8} = 1 \frac{1}{8} = 1.125 \\ x2 = \frac{1 - \sqrt{289} }{16} = \frac{1 - 17}{16} = \frac{ - 16}{16} = - 1$
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${x}^{4} - 16 {x}^{2} + 28 = 0 \\ {x}^{2} = y \\ {y}^{2} - 16y + 28 = 0 \\ d = {16}^{2} - 4 \times 28 = 256 - 112 = 144 \\ y1 = \frac{16 + \sqrt{144} }{2} = \frac{16 + 12}{2} = \frac{28}{2} = 14 \\ y2 = \frac{16 - \sqrt{144} }{2} = \frac{16 - 12}{2} = \frac{4}{2} = 2 \\ - - - - - - - - - - \\ {x}^{2} = y1 \\ {x}^{2} = 14 \\ x1 = \sqrt{14} \\ x3 = - \sqrt{14} \\ - - - - - - - - - - \\ {x}^{2} = y2 \\ {x}^{2} = 2 \\ x 2= \sqrt{2} \\x4 = - \sqrt{2}$