已知y=ax^5+bx^3+cx-5,其中a,b,c为常数,已知当x=-7时,y=7,求当x=7时,y的值等于

X=-7时,a*(-7)^5+b(-7)^3+c*(-7)-5=7
即-(a*7^5+b*7^3+c*7)=12
a*7^5+b*7^3+c*7=-12
x=7时有,y=a*7^5+b*7^3+c*7-5=-12-5=-17

f(7)=a(7)^5+b (7)^3+c*7-5
=-[a(-7)^5+b (-7)^3+c*(-7)]-5
=-[a(-7)^5+b (-7)^3+c*(-7)-5]-10
=-f(-7)-10
=-17

x=-7时，
y=-a*7^5-b*7^3-7c-5=-(a*7^5+b*7^3+7c)-5=7
所以a*7^5+b*7^3+7c=-12
x=7时，
y=a*7^5+b*7^3+7c-5
=-12-5
=-17

f(-7)=-7^5*a-7^3*b-7c-5=7
7^5*a+7^3*b+7c=-12
f(7)=7^5*a+7^3*b+7c-5=-17
所以当x=7时，y=-17

当x=-7时
y=a(-7)^5+b(-7)³+c(-7)-5
=-(a7^5+b7³+7c)-5
=7
得 -(a7^5+b7^3+7c)=12
所以 a7^5+b7^3+7c=-12
所以 x=7时
y=a7^5+b7^3+7c-5
=-12-5
=-17

y=f(x)=ax^5+bx^3+cx-5
那么f(-x)=-ax^5-bx^3-cx-5=-f(x)-10
故，f(7)=-f(-7)-10=-7-10=-17
有不懂欢迎追问