Решите систему уравнений: x^2-4xy+4y^2=25 x+2y=3

x^2-4xy+4y^2=25 x+2y=3  (х-2y)² = 5² x= 3-2y (x-2y)²-5² = 0 x = 3-2y (x-2y-5)*(x-2y+5) = 0 x = 3-2y (3-2y-2y-5)*(3-2y-2y+5) = 0 x = 3-2y (-4y-2)*(-4y+8) = 0 x= 3-2y -4y-2=0 -4y=2 y1=-1\2 -4y+8 = 0 -4y = -8 y2 = 2 x1= 3-2*(-1\2) = 4 x2 = 3-2*2= -1 Ответ: (4;-1\2) (-1;2)

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