Question

Решите уравнение
|x^2+2x-1|=(1-5x)/3

Answer 1

$| {x}^{2} + 2x - 1| = \frac{1 - 5x}{3}$

ОДЗ:

$\frac{1 - 5x}{3} ≥ 0 \\ 1-5x ≥ 0 \\ 5x ≤ 1 \\ x ≤ \frac{1}{5}$

Раскрываем модуль:

1)

${x}^{2} + 2x - 1 = \frac{1 - 5x}{3} \\ 3 {x}^{2} + 6x - 3 = 1 - 5x \\ 3 {x}^{2} + 11x - 4 = 0 \\ D = {11}^{2} - 4 \times 3( - 4) = 169 = {13}^{2} \\ x_{1} = \frac{ - 11 + 13}{6} = \frac{2}{6} = \frac{1}{3} \\ x_{2} = \frac{ - 11 - 13}{6} = \frac{ - 24}{6} = - 4$

2)

$x {}^{2} + 2x - 1 = - \frac{1 - 5x}{3} \\ 3{x}^{2} + 6x - 3 = 5x - 1 \\ 3 {x}^{2} + x - 2 = 0 \\ D = {1}^{2} - 4 \times 3( - 2) = 25 = {5}^{2} \\ x_{3} = \frac{ - 1 + 5}{6} = \frac{4}{6} = \frac{2}{3} \\ x_{4} = \frac{ - 1 - 5}{6} = - 1$

Проверяем какие корни вошли в ОДЗ:

$x_{1} = \frac{1}{3} \frac{1}{5} \Rightarrow x_{1} \in \varnothing \\ x_{2} = -4 < \frac{1}{5} \\ x_{3} = \frac{2}{3} \frac{1}{5} \Rightarrow x_{3} \in \varnothing \\ x_{4} = -1 < \frac{1}{5}$

ответ: -4; -1