若A{(X,Y)|Y=x2-2},B{(X,Y)|X=Y2-2},求A∩B

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若A{(X,Y)|Y=x2-2},B{(X,Y)|X=Y2-2},求A∩B若A{(X,Y)|Y=x2-2},B{(X,Y)|X=Y2-2},求A∩B若A{(X,Y)|Y=x2-2},B{(X,Y)|X

若A{(X,Y)|Y=x2-2},B{(X,Y)|X=Y2-2},求A∩B
若A{(X,Y)|Y=x2-2},B{(X,Y)|X=Y2-2},求A∩B

若A{(X,Y)|Y=x2-2},B{(X,Y)|X=Y2-2},求A∩B
求A∩B就是求Y=X²-2与X=Y²-2的公共解
两式相减得:Y-X=X²-Y²
故:(X-Y)(X+Y)+(X-Y)=0
故:(X-Y)(X+Y+1)=0
故:X=Y或X+Y+1=0
当X =Y时,代人Y=X²-2得:X=Y=2或X=Y=-1
当X+Y+1=0时,代人Y=X²-2得:X1=(-1+√5)/2,Y1=(-1-√5)/2或X2=(-1-√5)/2,Y2=(-1+√5)/2
故:A∩B={(2,2),(1,1),( (-1+√5)/2,(-1-√5)/2),( (-1-√5)/2,(-1+√5)/2)}