设f(x)是二次函数,且f(x-1)+f(2x+1)=5x^2+2x,求f(x)
问题描述：
设f(x)是二次函数,且f(x-1)+f(2x+1)=5x^2+2x,求f(x)
答
设f（X）=ax^2+bx+c
f(x-1)=a(x-1)^2+b(x-1)+c
f(2x+1)=a(2x+1)^2+b(2x+1)+c
f(x-1)+f(2x+1)=a(x-1)^2+b(x-1)+c+a(2x+1)^2+b(2x+1)+c
=ax^2-2ax+a+4ax^2+4ax+a+3bx+2c
=5ax^2+(2a+3b)x+2a+2c
=5x^2+2x
所以 5a=5
2a+3b=2
2a+2c=0
a=1 b=0 c=-1
f(x)=x^2-1