已知二次函数f（x）满足f（x+4）+f（x-1）=x²-2x,则,f（x）=

设二次函数：
f(x)=ax^2+bx+c
所以
f（x+4）+f（x-1）=a(x+4)^2+b(x+4）+c+a(x-1)^2+b(x-1）+c
=2ax^2+(6a+2b)x+17a+3b+2c=x²-2x
系数相等，自己解吧

f(x)=a(x-4)^2+b(x-4)+c,
f(x+4)=ax^2 + bx + c,
f(x-1)=a(x-5)^2 + b(x-5) + c = ax^2 - 10ax + 25a + bx - 5b + c,
x^2 - 2x = f(x+4)+f(x-1) = 2ax^2 + x(b-10+b) + 2c-5b,
1=2a, a = 1/2.
-2 = b-10+b, b = 4,
0 = 2c-5b, 2c=5b=20, c=10.
f(x)=(x-4)^2/2 + 4(x-4) + 10

设二次函数f（x）＝ax²＋bx＋c
则,f（x+4）+f（x-1）=2ax²＋(2b-6a)x＋(17a+3b+2c)=x²-2x
所以,
2a＝1
2b－6a＝－2
17a＋3b＋2c＝0
解得
a=1/2,b=1/2,c=-5
所以,
f（x）= x²/2＋x/2 -5