已知f(x)=(x^(1/3)-x^(-1/3))/5,g(x)=(x^(1/3)+x^(-1/3))/5

已知f(x)=(x^(1/3)-x^(-1/3))/5,g(x)=(x^(1/3)+x^(-1/3))/5

(1)f(x)=(x^(1/3)-x^(-1/3))/5
g(x)=(x^(1/3)+x^(-1/3))/5
X属于R
则有:f(-x)=[(-x)^(1/3)-(-x)^(-1/3)]/5
=[x^(-1/3)-x^(1/3)]/5
=-[x^(1/3)-x^(-1/3)]/5
=-f(x)
g(-x)=[(-x)^(1/3)+(-x)^(-1/3)]/5
=[-x^(1/3)-x^(-1/3)]/5
=-[x^(1/3)+x^(-1/3)]/5
=-g(x)
设Y=T(X)=f(x)*g(x)
则有:T(-x)=f(-x)*g(-x)
=[-f(x)]*[-g(x)]
=f(x)g(x)
=T(x)
则y=f(x)*g(x)为偶函数
则有T(X)=f(x)*g(x)图像关于Y轴对称
又y=f(x)*g(x)
={[x^(1/3)-x^(-1/3)]/5}*{[x^(1/3)+x^(-1/3)]/5}
=[x^(2/3)-x^(-2/3)]/5
则任取X1,X2属于(0,+无穷)
且X1>X2
则有:X1^(1/3)>X2^(1/3)>0
0 则:f(x1)-f(x2)
=[x1^(2/3)-x1^(-2/3)]/5-{[x2^(2/3)-x2^(-2/3)]/5}
={[x1^(2/3)-x2^(2/3)]-[x1^(-2/3)-x2^(-2/3)]}/5
={[x1^(1/3)+x2^(1/3)]*[x1^(1/3)-x2^(1/3)]-
[x1^(-1/3)+x2^(-1/3)]*[x1^(-1/3)-x2^(-1/3)]}/5
又X1^(1/3)>X2^(1/3)>0
0则有: x1^(1/3)+x2^(1/3)>0
x1^(1/3)-x2^(1/3)>0
x1^(-1/3)+x2^(-1/3)>0
x1^(-1/3)-x2^(-1/3)<0
则:[x1^(1/3)+x2^(1/3)]*[x1^(1/3)-x2^(1/3)]>0,
[x1^(-1/3)+x2^(-1/3)]*[x1^(-1/3)-x2^(-1/3)]<0
则f(x1)-f(x2)
={[x1^(1/3)+x2^(1/3)]*[x1^(1/3)-x2^(1/3)]-
[x1^(-1/3)+x2^(-1/3)]*[x1^(-1/3)-x2^(-1/3)]}/5>0
即:X1>X2>0时,f(x1)>f(x2)
则T(X)在(0,+无穷)上单调递增
又y=T(X)=f(x)*g(x)为偶函数
则有T(X)在(-无穷,0)上单调递减

(2)f(4)-5f(2)*g(2)
=[4^(1/3)-4^(-1/3)]/5-5*{[2^(1/3)-2^(-1/3)]/5}*{[2^(1/3)+2^(-1/3)]/5}
=[4^(1/3)-4^(-1/3)]/5-{[2^(1/3)-2^(-1/3)]*[2^(1/3)+2^(-1/3)]/5}
=[4^(1/3)-4^(-1/3)]/5-[4^(1/3)-4^(-1/3)]/5
=0
同理可得:f(9)-5f(3)*g(3)=0
归纳得：f(x^2)－5f(x)g(x)＝0

这个题目简单。。自己看看书