1По определению модуля:
x³ - 3x² + x = 2x - x² или x³ - 3x² + x = - ( 2x - x²)
x³ - 3x² + x² + x - 2x = 0 x³ - 3x² + x = - 2x + x²
x³ - 2x² - x = 0 x³ - 3x² - x² + x + 2x = 0
x( x² - 2x - 1)=0 x³ - 4x² + 3x=0
x1= 0 x(x² - 4x + 3) = 0
x² - 2x - 1 = 0 x² - 4x + 3 = 0
D = b² - 4ac = 4 - 4 × (-1)=8 D = b²-4ac = 16 - 4×3 = 16-12 = 4 = 2²
x2 =( 2 + √8) / 2 = (2 + 2√2) / 2 = x4 = ( 4 + 2) / 2 = 3
= 2( 1 + √2)/2 = 1 + √2 x5 = ( 4 - 2)/2 = 1
x3 = (2 - √8)/2 = 1 - √2
Ответ: имеет пять корней:x1 = 0, x2 = 1 + √2, x3 = 1 - √2, x4 = 3, x5 = 1.