Question

Найдите корни дробно-рациональных уравнений:
1)
$\frac{ {x}^{2} - 6x } {3x - 1} = \frac{ {3x }^{2} - 4x}{1 - 3x}$
2)
$\frac{y + 4}{y - 3} = \frac{2y - 2}{y}$
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Answer 1

В решении

Объяснение:

$1)\frac{ {x}^{2} - 6x }{3x - 1} = \frac{3 {x}^{2} - 4x}{1 - 3x} \\ \frac{ {x}^{2} - 6x }{3x - 1} = \frac{3 {x}^{2} - 4x}{ - (3x - 1)} \\ {x}^{2} - 6x = - 3 {x}^{2} + 4x \\ 4 {x}^{2} - 10x = 0 \\ 2x(2x - 5) = 0 \\ x = 0 \: x = 2.5 \\ 2) \frac{y + 4}{y - 3} = \frac{2y - 2}{y} \\ (2y - 2)(y - 3) = y(y + 4) \\ 2 {y}^{2} - 6y - 2y + 6 = {y}^{2} + 4y \\ {y}^{2} - 12y + 6 = 0 \\y = \frac{6 + \sqrt{36 - 6} }{1} = 6 + \sqrt{30} \\ y = \frac{6 - \sqrt{36 - 6} }{1} = 6 - \sqrt{30}$