已知3f(x^5)+f(-x^5)=4x,求f(x)

令x=-x
则f(x^5)=f(-x^5)
f(-x^5)=f(x^5)
所以3f(-x^5)+f(x^5)=-4x (2)
3f(x^5)+f(-x^5)=4x (1)
(1)*3-(2)
9f(x^5)-f(x^5)=16x
f(x^5)=2x
另a=x^5
x=a^(1/5)
f(a)=2a^(1/5)

所以f(x)=2x^(1/5)

已知3f(x^5)+f(-x^5)=4x，所以3f(x)+f(-x)=4x^(1/5),
所以3f(-x)+f(x)=4(-x)^(1/5),
1式×3-2式，得f(x)=[3x^(1/5)]/2-[(-x)^(1/5)]/2.

令t=-x
则3f(-t^5)+f(t^5)=-4t,即3f(-x^5)+f(x^5)=-4x
两式计算可得f(x^5)=2x
所以f(x)=(我用文字表达吧) 2倍的 五次根号下X

3f(x^5) + f(-x^5) = 4x
取-x带入，可知，
3f(-x^5) + f(x^5) = -4x
解之，可得f(x^5) = 2x
在进行变换，可得，f(x)=2*五次根号下x

3f(x^5)+f(-x^5)=4x
1.若为奇函数，有3f(x^5)-f(x^5)=4x：f(x)=（2x）^(1/5);
2.若为偶函数，则3f(x^5)+f(x^5)=4x：f(x)=x^(1/5),与他为奇函数矛盾.