1.设A={x|x2-3x+2=0} B={x|x2-ax+a-1=0} C={x|x2-bx+2=0}若A∪B=A,A

2个回答

  • 因为(自己改成符号)集合A={x|x2-3x+2=0},所以A={1,2}(因式分解法)

    因为A∪B=A,A∩C=C,

    所以B含于A,C含于A

    所以可将X=2或X=1代入x2-ax+a-1=0,x2-bx+2=0

    得a=3,b=1或b=3

    2、由题意得X2+PX1+Q=0 X2+PX2+Q=0

    (X1+1)2+Q(X1+1)+P=0=X12+2X1+1+QX1+X1+P+Q

    (X2+1)2+Q(X2+1)+P=0=X22+2X2+1+QX2+Q+P

    X12+2X1+1+QX1+X1+P+Q=X2+PX1+Q

    PX1=2X1+1+QX1+X1+P

    X22+2X2+1+QX2+Q+P=X2+PX2+Q

    PX2=2X2+1+QX2+P

    所以 PX1-PX2=2X1-2X2

    P=2 所以 X12+2X1+Q=0

    X22+2X2+Q=0

    则X12+2X1=X22+2X2

    X12-X22=2X2-2X1

    (X1+X2)(X1-X2)=2(X2-X1)

    -X1-X2=-2

    因为(X1+1)2+QX1+2=0=X12+2X1+QX1+2

    (X2+1)2+QX2+2=0=X22+2X2+1+QX2

    所以X12+QX1=X22+QX2

    X12-X22=QX2-QX1

    (X1+X2)(X1-X2)=Q(X2-X1)

    Q=-X1-X2=-2

    综上所述P=2 Q=-2