Ответы на вопрос:
2х(х-4)(х+5)=0 2х=0 х1=0 х-4=0 х2=4 х+5=0 х3=-5 х^2-25=0 х^2=25 х=+-! /25 х1=-5 х2=5 х^2-8х+16=0 д=64-4×1×16=0 х=8/2=4
{y=x^2 {x-y=0x - x² = 0 x(1-x) =0 x = 0 1-x =0 -x = -1 x = 1 y = 0 y = 1^2 = 12) ( y-2)^2-4 = (y-2)² - 2² = (y-2 -2)(y-2 +2) = (y-4)y x^2+2xy+y^2 = (x +y)²
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