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Okesan
@Okesan
August 2022
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Найдите все пары целых решений уравнения
2x^2+y^2=2xy+4x
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konik1
2x^2 + y^2 = 2xy + 4x
2x^2 + y^2 - 2xy = 4x
(x - y)^2 = 4x - x^2
(x - y)^2 = x(4 -x)
x - y = 0 4 - x = 0
y =4
x = 4
(x;y) = (4;4)
x - y = √ x(4 -x)
x(4 -x) = x^2
4 - x = x
2x = 4
x=2
2*
2^2 + y^2 - 2*
2y = 4*
2
8 +
y^2 -4y = 8
y(y-4) = 0
y =4
(x;y)
= (2;4)
2 votes
Thanks 1
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Answers & Comments
2x^2 + y^2 - 2xy = 4x
(x - y)^2 = 4x - x^2
(x - y)^2 = x(4 -x)
x - y = 0 4 - x = 0
y =4 x = 4 (x;y) = (4;4)
x - y = √ x(4 -x)
x(4 -x) = x^2
4 - x = x
2x = 4
x=2
2*2^2 + y^2 - 2*2y = 4*2
8 + y^2 -4y = 8
y(y-4) = 0
y =4 (x;y) = (2;4)