$(\frac{2x - 1}{x})^2 - \frac{6(2x - 1)}{x} + 5 = 0$
$y = \frac{2x - 1}{x}$
$y^2 - 6y + 5 = 0$
$D = b^2 - 4ac = (-6)^2 - 4 * 1 * 5 = 36 - 20 = 16 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{6 + \sqrt{16}}{2 * 1} = \frac{6 + 4}{2} = \frac{10}{2} = 5$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{6 - \sqrt{16}}{2 * 1} = \frac{6 - 4}{2} = \frac{2}{2} = 1$
$\frac{2x - 1}{x} = 5$
2x − 1 = 5x
2x − 5x = 1
−3x = 1
$x = -\frac{1}{3}$
или
$\frac{2x - 1}{x} = 1$
2x − 1 = x
2x − x = 1
x = 1
Ответ: $-\frac{1}{3}$ и 1

$\frac{3x - 1}{x + 1} + \frac{x + 1}{3x - 1} = 3\frac{1}{3}$
$\frac{3x - 1}{x + 1} + \frac{x + 1}{3x - 1} = \frac{10}{3}$
x + 1 ≠ 0
x ≠ −1
и
3x − 1 ≠ 0
3x ≠ 1
$x ≠ \frac{1}{3}$
$y = \frac{3x - 1}{x + 1}$
$y + \frac{1}{y} = \frac{10}{3}$
y ≠ 0
$y + \frac{1}{y} = \frac{10}{3}$ | * 3y
$3y^2 + 3 = 10y$
$3y^2 - 10y + 3 = 0$
$D = b^2 - 4ac = (-10)^2 - 4 * 3 * 3 = 100 - 36 = 64 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{10 + \sqrt{64}}{2 * 3} = \frac{10 + 8}{6} = \frac{18}{6} = 3$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{10 - \sqrt{64}}{2 * 3} = \frac{10 - 8}{6} = \frac{2}{6} = \frac{1}{3}$
$\frac{3x - 1}{x + 1} = 3$ | * (x + 1)
3x − 1 = 3(x + 1)
3x − 1 = 3x + 3
3x − 3x = 3 + 1
0 ≠ 4
нет корней
$\frac{3x - 1}{x + 1} = \frac{1}{3}$ | * 3(x + 1)
3(3x − 1) = x + 1
9x − 3 = x + 1
9x − x = 1 + 3
8x = 4
x = 0,5
Ответ: 0,5