55 Given the function $f(x)$ satisfies for all real numbers $x, y$, 
$$
f(x)+f(2 x+y)+5 x y=f(3 x-y)+2 x^{2}+1,
$$

then the value of $f(10)$ is ( ).
(A) -49
(B) -1
(C) 0
(D) 25
(5) Let $x=10, y=5$, we get
$$
f(10)+f(25)+250=f(25)+200+1 \text {, }
$$

so $f(10)=-49$. Therefore, the answer is A.