Question

帮忙解答一道初一数学的题目!!谢谢大家

当x=-3时,代数式ax^5+bx^3+cx-8=6,求当x=3时,代数式ax^5+bx^3+cx-8的值

Answer 1

当x=-3时含铅昌，a（-3）^5+b（-3）^3+c（-3）-8=6
=>-（a3^5+b3^3+c3）=8+6=14
=>a3^5+b3^3+c3=-14
所以，激敏当谈扒x=3时，
ax^5+bx^3+cx-8
=a3^5+b3^3+c3-8
=-14-8=-22

代数式ax^5+bx^3+cx-8的值是-22

Answer 2

当x=-3时，a（-3）^5+b（-3）^3+c（贺樱-3）-8=6
=>-（a3^5+b3^3+c3）=8+6=14
=>a3^5+b3^3+c3=-14
所以瞎模，当x=3时磨拍缓，
ax^5+bx^3+cx-8
=a3^5+b3^3+c3-8
=-14-8=-22

Answer 3

x为负数时 它的奇旁李数判誉次幂与偶数次幂互为相运冲迟反数 x=-3 ax^5+bx^3+cx=14 所以x=3 ax^5+bx^3+cx=-14 ax^5+bx^3+cx-8=-22

Answer 4

5 3 1是单数型纳雹，所以当x=-3时的ax^5+bx^3+cx=-（当x=3时茄没ax^5+bx^3+cx
ax^5+bx^3+cx-8=6则当x=-3时的ax^5+bx^3+cx=14
当卜帆x=3时ax^5+bx^3+cx=-14
-14-8=22