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Реши систему уравнений: {y+x=−1 x−y=12
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Ответ:

) x+y=12 _ x=12-y _ x=12-y

xy=32 (12-y)y=32 12y-y^2=32

y^2-12y+32=0

D=144-4*32=16

y1=(12+4)/2=8 x1=12-8=4

y2=(12-4)/2=4 x2=12-4=8

б) y=x+2 _ y=x+2 _ y=x+2

4y+x^2=8 4(x+2)+x^2=8 4x+8+x^2=8

x^2+4x=0

x(x+4)=0

x1=0 y1=2

x2=-4 y2=-4+2=-2

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