solve x
(1+x)^5 = 15x
數學題 (1+x)^5 = 15x
2009-04-27 9:13 pm
回答 (8)
2009-04-27 10:39 pm
✔ 最佳答案
We use the Newton's method to find the approximate root as follows:Let f(x) = (1 + x)5 - 15x, then we have:
f(0) = 1 and f(1) = -15
Thus, we are sure that there's a root between x = 0 and x = 1
Also f'(x) = 5(1 + x)4 - 15
Starting from x0 = 0:
x1 = x0 - f(x0)/f'(x0) = 0.1
x2 = x1 - f(x1)/f'(x1) = 0.11439
x3 = x2 - f(x2)/f'(x2) = 0.11477
x4 = x3 - f(x3)/f'(x3) = 0.11477
Thus, correcting to 5 d.p., we can approximate x = 0.11477
參考: Myself
2009-04-28 11:26 pm
It should have 3 distinct real roots.
2009-04-28 11:33 am
D人係咪傻o架?
好明顯就唔係 1/2 lor..
好明顯就唔係 1/2 lor..
2009-04-28 4:47 am
5+5x=15x
5=10x
x=1/2
5=10x
x=1/2
2009-04-27 10:08 pm
(1+x)^5=15x
(1+x)(1+x)(1+x)(1+x)(1+x)=15x
(1+2x+x^2)(1+x)(1+x)(1+x)=15x
(1+3x+3x^2+x^3)(1+x)(1+x)=15x
(1+4x+6x^2+4x^3+x^4)(1+x)=15x
(1+5x+10x^2+10x^3+5x^4+x^5)=15x
...............Can't solve x...............
(1+x)(1+x)(1+x)(1+x)(1+x)=15x
(1+2x+x^2)(1+x)(1+x)(1+x)=15x
(1+3x+3x^2+x^3)(1+x)(1+x)=15x
(1+4x+6x^2+4x^3+x^4)(1+x)=15x
(1+5x+10x^2+10x^3+5x^4+x^5)=15x
...............Can't solve x...............
參考: me
2009-04-27 9:19 pm
5+5x=15x
5=10x
x=1/2(最後答案)
5=10x
x=1/2(最後答案)
收錄日期: 2021-04-13 16:35:56
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