f'(0)=1,f(x+y)=f(x)+f(y)+2xy,求f(x)

5xy²-[3xy²-(4xy²-2x²y)]+2x²y-xy²
=5xy²-3xy²+(4xy²-2x²y)+2x²y-xy²
=5xy²-3xy²+4xy²-2x²y+2x²y-xy²
=(5-3+4-1)xy²+(-2+2)x²y
=5xy²

游客1:定义在R上的函数f(x)满足f(x+y)=f(x)+f(y)+2xy(x,y∈R)f(1)=2,
则f(-3)=?
2:设f(x)是连续偶函数,且当x>0时f(x)是单调函数,则满足f(x)=f(x+3/x+4)的所有x之和为?
3:函数f(x)满足:⑴定义域是(0,+∞)⑵当x>1时,f(x)

假设y > 0,记y = Δx,则有
f(x + Δx) - f(x) = f(Δx) + 2xΔx
从而有
[f(x + Δx) - f(x)]/Δx = f(Δx) + 2x
对上式取极限,使得Δx趋近于0+,则左式就是f'(x),即
f'(x) = limf(Δx) + lim2x =f(0) + 2x
所以现在要把f(0)搞定,令x = 0,有
f'(0) = f(0) = 2
所以
f'(x) = 2x + 2
f(x) = x^2 + 2x + C
代入f(0) = 2,有C = 2
所以f(x) = x^2 + 2x + 2

1.f（1）=f(1)+f(0)+0,故f(0)=0 f(2)=f(1)+f(1)+2*1*1=6 f(3)=f(1)+f(2)+2*1*2=12 又f(0)=f(x)+f(-x)-2x^2=0 f(-3)=2*3^2-f(3)=6.