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AnnaDor
@AnnaDor
August 2022
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Решите систему уравнений методом подстановки
x•y= -2
x+y=1
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zaika2809
Ху = - 2
х + у= 1 → х = 1 - у, подставляем значение х в 1-ое уравнение,
получаем:
у(1-у) = - 2
у - у^2 = -2
-y^2 + y + 2 = 0
y^2 - y - 2 = 0
D = 1 - 4 * -2 = 1+8 = 9 √D = 3
y1 = (1+3)/2 = 2
y2 = (1-3)/2 = - 1
Подставляем найденное значение у1 и у2 во 2-ое уравнение:
х + у = 1 х + у = 1
х + 2 = 1 х - 1 = 1
х = -2 + 1 х = 1 + 1
х1 = - 1 х2 = 2
Ответ: х1 = - 1 х2 = 2
у1 = 2 у2 = - 1
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кляча
Verified answer
X=1-y (1-y)×y=-2
y-y^2=-2
-y^2+y+2=0 (-1)
y^2-y-2=0
D=1^2+2×8=1+8=9=3^2
y1=(1+3)/2=4/2=2
y2=(1-3)/2=-2/2=-1
x1=1-2=-1
x2=1-(-1)=1+1=2
(2;-1) и (-1;2)
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Answers & Comments
х + у= 1 → х = 1 - у, подставляем значение х в 1-ое уравнение,
получаем:
у(1-у) = - 2
у - у^2 = -2
-y^2 + y + 2 = 0
y^2 - y - 2 = 0
D = 1 - 4 * -2 = 1+8 = 9 √D = 3
y1 = (1+3)/2 = 2
y2 = (1-3)/2 = - 1
Подставляем найденное значение у1 и у2 во 2-ое уравнение:
х + у = 1 х + у = 1
х + 2 = 1 х - 1 = 1
х = -2 + 1 х = 1 + 1
х1 = - 1 х2 = 2
Ответ: х1 = - 1 х2 = 2
у1 = 2 у2 = - 1
Verified answer
X=1-y (1-y)×y=-2y-y^2=-2
-y^2+y+2=0 (-1)
y^2-y-2=0
D=1^2+2×8=1+8=9=3^2
y1=(1+3)/2=4/2=2
y2=(1-3)/2=-2/2=-1
x1=1-2=-1
x2=1-(-1)=1+1=2
(2;-1) и (-1;2)