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