Question

рівняння
$x {}^{4} - 29x { }^{2} + 100 = 0$
рівняння
$\frac{3x - 1}{x + 1} = \frac{ x - 1}{x - 2}$
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Answer 1

$x {}^{4} - 29x { }^{2} + 100 = 0 \\ {x}^{2} = t \\ {t}^{2} - 29t + 100 = 0 \\D = {b}^{2} - 4ac = {29}^{2} - 4 \times 1 \times 100 = 441 \\ t_{1} = \frac{ - b - \sqrt{D} }{2a} = \frac{29 - \sqrt{441} }{2 \times1 } = \frac{8}{2} = 4 \\ t_{2} = \frac{ - b + \sqrt{D} }{2a} = \frac{29 + \sqrt{441} }{2 \times 1} = \frac{50}{2} = 25 \\ t_{1} = x_{ \\1 ,2 {}}^{2} \\ x_{ \\1 ,2 {}} = \sqrt{4} \\ x_{ \\1 ,2 {}} = ±2 \\ t_{2} = x_{ \\3 ,4{}} \\ x_{ \\3 ,4{}} = \sqrt{25 \\ } \\ x_{ \\3 ,4{}} =±5$

$\frac{3x - 1}{x + 1} = \frac{ x - 1}{x - 2} \\ (3x - 1)(x - 2) = (x + 1)(x - 1) \\ 3 {x}^{2} - x - 6x + 3 = {x}^{2} - 1 \\ 2 {x}^{2} - 7x + 4 = 0 \\ D = {b}^{2} - 4ac =( - 7 {)}^{2} - 4 \times 2 \times 4 = 17 \\ x_{1} = \frac{ - b - \sqrt{D} }{2a} = \frac{7 - \sqrt{17} }{2 \times 2} = \frac{7 - \sqrt{17} }{4} \\ x_{2} = \frac{ - b + \sqrt{D} }{2a} = \frac{7 + \sqrt{17} }{2 \times 2} = \frac{7 + \sqrt{17} }{4}$