Question

y=ax5的平方+bx5的平方+cx减5,期中abc是常数,当x=-5时,y=7,求当x=5时,球y的值

Answer 1

当x=-5时,y=7

即a*(-5)^5+b*(-5)^5+c*(-5)-5=7

所以a*5^5+b*5^5+5c+5=-7

a*5^5+b*5^5+5c=-12

当x=5时

y=a*5^5+b*5^5+5c-5

=-12-5

=-17