數學 疊加法求公式解 幫幫忙?
回答 (4)
2017-03-26 3:33 am
✔ 最佳答案
b + c+ d+ e = -c(5,1)*2^4*3 + c(5,2)*2^3*3^2 - c(5,3)2^2*3^3 + c(5,4)2*3^4= -240 +720 - 1080 + 810
= 1530 - 1320
= 210
= Answer
2017-03-26 9:16 pm
a=2^5=32
x=0 則f=(-3)^5=-243
x=1 則-1= a+b+c+d+e+f
b+c+d+e=210
x=0 則f=(-3)^5=-243
x=1 則-1= a+b+c+d+e+f
b+c+d+e=210
2017-03-26 8:18 am
若(2x−3)^5 =ax^5+bx^4+cx^3+dx^2+ex+ f ，求b+c+d+e的值為?
b+c+d+e = (a+b+c+d+e+f) - (a+f)
(2*1-3)^5 = a+b+c+d+e+f = -1
(2x-3)^5 = 2^5 x5 +... + (-3)^5
a = 2^5 = 32, f = (-3)^5 = -243
故 b+c+d+e = (-1) - [32+(-243)] = 210
b+c+d+e = (a+b+c+d+e+f) - (a+f)
(2*1-3)^5 = a+b+c+d+e+f = -1
(2x-3)^5 = 2^5 x5 +... + (-3)^5
a = 2^5 = 32, f = (-3)^5 = -243
故 b+c+d+e = (-1) - [32+(-243)] = 210
2017-03-25 11:23 pm
(2X-3)^5=C(5,0)(2X)^5+C(5,1)(2X)^4(-3)+C(5,2)(2X)^3(-3)^2+C(5,3)(2X)^2(-3)^3+C(5,4)(2X)(-3)^4+C(5,5)(-3)^5=32X^5-240X^4+720X^3-1080X^2+810X-243
所以B+C+D+E=-240+720-1080+810-243=-13
所以B+C+D+E=-240+720-1080+810-243=-13
收錄日期: 2021-05-01 15:28:01
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20170325141426AA6NdV6