Найди корни уравнения √6x2−2x+1=√3x+2
ответ:
x1=
x2=−

√6x² - 2x + 1 = √3x + 2

6x² - 2x + 1 = 3x + 2

6x² - 2x + 1 - 3x - 2 = 0

6x² - 5x - 1 = 0

6x² + x - 6x - 1 = 0

x(6x + 1) - (6x + 1) = 0

(6x + 1)(x - 1) = 0

6x + 1 = 0

6x = -1

x₁ = -1/6

x - 1 = 0

x₂ = 1

x₁ = -1/6

x₂ = 1