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x²+y²=29
y=3-x
x²+(3-x)²=29
x²+9-6x+x²=29
2x²-6x-20=0 /;2
x²-3x-10=0
Δ=b²-4ac=9+40=49
x1=[3-7]/2=-2 y1=3+2=5
x2=[3+7]/2=5 y2=3-5=-2
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A; (x-y)²=(-2-5)²=49 lub ( 5+2)²=49 = fałsz
B; xy=4 -2*5=-10 lub 5*(-2)=-10 = fałsz
C;(x+y)²=[-2+5)²=9 lub ( 5-2]²=9
odp. D