x^3-2x+1=0
Найдите (2-a)(2-b)(2-c), где a,b,c корни кубического уравнения

x^3 - 2x + 1 = 0;0=x^3 - x - x +1 = x*( x^2 - 1) - (x-1) = x*(x-1)*(x+1) - (x-1) = = (x-1)*( x*(x+1) - 1) = (x-1)*( x^2 + x - 1) = 0;1) x-1 = 0, <=> x_1=1или2) x^2 + x - 1 = 0;D = 1 - 4*(-1) = 1+4 = 5.x_2= (-1 -√5)/2;x_3 = (-1+√5)/2;(2-a)(2-b)(2-c) = (2 - x_1)*(2 - x_2)*(2 - x_3) = = (2-1)*(2 - [(-1-√5)/2] )*(2 - [(-1+√5)/2] ) == 1*(4 + 1 + √5)*(4 +1 -√5)/4 = (5+√5)*(5-√5)/4 = (25 - 5)/4 = 20/4 = 5.