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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