Система уравнений 2x^2-3xy+y^2=0 y^2-x^2=12 - вопрос №8173526223202212 от pampuzik 15.09.2022 00:36

Question

Математика: Система уравнений 2x^2-3xy+y^2=0 y^2-x^2=12

pampuzik · Answer

{x² - xy + (x-y)² = 0{(y-x)(x+y) = 12{-x(y-x) + (y-x)² = 0{(y-x)(x+y) = 12{(y-x-x)(y-x) = 0{(y-x)(y+x) = 12y≠x y = 2x(2x-x)(2x+x) = 123x² = 12x² = 4x1 = -2x2 = 2y1 = -4y2 = 4ответ: (-2; -4); (2; 4)...