Question

найдите корень уравнения плиииз
$\frac{2x + 1}{3} = x \: \: \: \: \: \frac{ {x}^{2} - 9x } {10} = \frac{1 - 4x}{5}$
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Answer 1

1)

$\frac{2x + 1}{ 3} - x = 0 \\ \frac{2x + 1 - 3x}{3} = 0 \\ 2x - 3x + 1 = 0 \\ - x + 1 = 0 \\ - x = - 1 \\ x = 1$

2)

$\frac{ {x}^{2} - 9x }{10} - \frac{1 - 4x}{5} = 0 \\ \frac{ {x}^{2} - 9x - 2(1 - 4x) }{10} = 0 \\ {x}^{2} - 9x - 2 + 8x = 0 \\ {x}^{2} - x - 2 = 0 \\ D = {( - 1)}^{2} - 4 \times 1 \times ( - 2) = 1 + 8 = 9 \\ \sqrt{D} = \sqrt{9} = 3 \\ x1 = \frac{1 + 3}{2} = \frac{4}{2} = 2 \\ x2 = \frac{1 - 3}{2} = \frac{ - 2}{2} = - 1$