Решите уравнения: ж) (5y+1)/(y+1) = (y+2)/y; з) (1+3x)/(1-2x) = (5-3x)/(1+2x); и) (x-1)/(2x+3) - (2x-1)/(3-2x) = 0

ж) $\frac{5y+1}{y+1} = \frac{y+2}{y}$ ОДЗ: $y \neq -1; y \neq 0$. $y(5y+1) = (y+1)(y+2)$ $5y^2 + y = y^2 + 2y + y + 2$ $5y^2 + y = y^2 + 3y + 2$ $4y^2 - 2y - 2 = 0$ | $: 2$ $2y^2 - y - 1 = 0$ $D = (-1)^2 - 4 \cdot 2 \cdot (-1) = 1 + 8 = 9$ $y_1 = \frac{1+3}{4} = 1$; $y_2 = \frac{1-3}{4} = -0,5$ Ответ: -0,5; 1 з) $\frac{1+3x}{1-2x} = \frac{5-3x}{1+2x}$ ОДЗ: $x \neq 0,5; x \neq -0,5$. $(1+3x)(1+2x) = (5-3x)(1-2x)$ $1 + 2x + 3x + 6x^2 = 5 - 10x - 3x + 6x^2$ $1 + 5x = 5 - 13x$ $18x = 4$ $x = \frac{4}{18} = \frac{2}{9}$ Ответ: \frac{2}{9} и) $\frac{x-1}{2x+3} - \frac{2x-1}{3-2x} = 0$ ОДЗ: $x \neq -1,5; x \neq 1,5$. $\frac{x-1}{2x+3} + \frac{2x-1}{2x-3} = 0$ $(x-1)(2x-3) + (2x-1)(2x+3) = 0$ $2x^2 - 3x - 2x + 3 + 4x^2 + 6x - 2x - 3 = 0$ $6x^2 - x = 0$ $x(6x - 1) = 0$ $x_1 = 0$; $x_2 = \frac{1}{6}$ Ответ: 0; \frac{1}{6}