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LG2002123
@LG2002123
July 2022
1
8
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пусть (Х0; У0) - решение системы х- 2у= -3
у квадрат -2х=3.
найдите Х0+2У0
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miron2077
Verified answer
X - 2y = - 3
y^2 - 2x = 3
••••••
X = 2y - 3
2x = 4y - 6
2x = y^2 - 3
••••••••
4y - 6 = y^2 - 3
y^2 - 4y + 3 = 0
D = 16 - 12 = 4 = 2^2
y1 = ( 4 + 2 ) : 2 = 3
y2 = ( 4 - 2 ) : 2 = 1
••••••••••
X = 2y - 3
X1 = 6 - 3 = 3
X2 = 2 - 3 = - 1
____________
1) 3 + 2*3 = 9
2) - 1 + 2*1 = 1
Ответ 9; 1
4 votes
Thanks 12
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Answers & Comments
Verified answer
X - 2y = - 3y^2 - 2x = 3
••••••
X = 2y - 3
2x = 4y - 6
2x = y^2 - 3
••••••••
4y - 6 = y^2 - 3
y^2 - 4y + 3 = 0
D = 16 - 12 = 4 = 2^2
y1 = ( 4 + 2 ) : 2 = 3
y2 = ( 4 - 2 ) : 2 = 1
••••••••••
X = 2y - 3
X1 = 6 - 3 = 3
X2 = 2 - 3 = - 1
____________
1) 3 + 2*3 = 9
2) - 1 + 2*1 = 1
Ответ 9; 1