Question

设f(x)=X2+bx+c,集合A={x|f（x）=x}，B={x|f（x-1）=x+1，若集合A={2}，求集合B

x2是指x的平方

Answer 1

xx+(b-1)x+c=0
只有解2
b-1=-4 => b=-3
c=4

f(x-1）=(x-1)^2-3(x-1)+4=x+1
=>xx-6x+7=0
x=3+√2或3-√2
B={3+√2，3-√2}