第1个回答 2008-09-05
集合A={2}
x2+px+q=x,即:x2+(p-1)x+q=0有惟一解:2
△=(p-1)^2-4q=0
q=(p-1)^2/4
把:x=2代入:x2+(p-1)x+(p-1)^2/4=0得:
4+2(p-1)+(p-1)^2/4=0
(p-1)^2+8(p-1)+16=0
[(p-1)+4]^2=0
(p+3)^2=0
p=-3
q=(p-1)^2/4=4
f(x-1)=x+1
(x-1)^2+p(x-1)+q=x+1
(x-1)^2-3(x-1)+4=x+1
x^2-6x+7=0
x=3±√2
B={3+√2,3-√2}
第2个回答 2008-09-05
A={x|f(x)=x,x属于R}={2}
f(x)=x^2+px+q
所以
x^2+(p-1)x+q=0 ==> x=2 (重根..)
所以
q=4, p-1=-4
p=-3
所以
x|f(x-1)=x+1
(x-1)^2+p(x-1)+q = x+1
(x-1)^2 - 3(x-1) - x - 1 + 4 = 0
(x-1)^2 - 4(x-1) + 2 = 0
x1-1 = 2 + sqrt(2), x2-1 = 2 - sqrt(2)
x1 = 3 + sqrt(2)
x2 = 3 - sqrt(2)
所以
B = {3 + 根号2, 3 - 根号2}
第3个回答 2008-09-05
A={2}==>x^2+(p-1)x+q=0的解为x1=x2=2 ==>p=-3,q=4
f(x-1)=x+1 ==> (x-1)^2-3(x-1)+4=x+1 ==>x^2-6x+7=0
==>...
第4个回答 2008-09-05
(1)x2+px+q=x
(2)x=2
结论:2p+q=-2
(1):(x-1)2+p(x-1)+q=x+1
(2):x=2
结论:p+q=2
结论:p=-4 q=6
代入 (x-1)2+p(x-1)+q=x+1
得 : x2-7x+10=0
算得 x=2 或 x=5
结论:B=【2,5】