Question

数学问题，急用

已知 ax的5次方+bx的3次方+cx+3 ,当x=3时 它的值为48，求，当x为 负3 时，它的值为多少

过程，谢咯

Answer 1

解：ax5+bx3+cx+3
当x=3时原式=a(3)5+b(3)3+c(3)+3=48所以a(3)5+b(3)3+3c=45
当x=-3时原式=a(-3)5+b(-3)3+c(-3)+3
=-a(3)5-b(3)3-3c+3
=-【a(3)5+b(3)3+3c】+3
=-45+3=-42

Answer 2

x=3时
aX^5+bX^3+cX+d
=a×3^5+b×3^3+c×3+d
=-4

x=-3时
aX^5+bX^3+cX-d
=a×(-3)^5+b×(-3)^3+c×(-3)-d
=-a×3^5-b×3^3-c×3-d
=-(a×3^5+b×3^3+c×3+d)
=-(-4)
=4

Answer 3

解：因为3^5a=-(-3^5a) 3^3b=-(-3^3b) 3c=-(-3c)
所以 当x=-3时原式=-（3^5a+3^3b+3c)+3
=-(3^5a+3^3b+3c+3-3)+3
=-(48-3)+3
=-45+3
=-42
注：“^”符号是次方的意思

Answer 4

3^5a=-(-3^5a) 3^3b=-(-3^3b) 3c=-(-3c)
当x=-3时原式=-（3^5a+3^3b+3c)+3
=-(3^5a+3^3b+3c+3-3)+3
=-(48-3)+3
=-45+3
=-42