因不能打二次方的符號, 所以把所有的二次方打成[square]
三次方--> [cube]
求steps!!!!!
Factorization因式分解
1)(x square+3x)square+3(x square+3x)+2
ans= (x+1)(x+2)(x square+3x+1)
2) 2x cube - 3x square +1
ans= (x-1)square(2x+1)
Formulae公式
1)It is known that the sum of the first n positive integers can be found by the following formula: S= n(n+1)/2
q= find the sum of the positive integers from 11to50
ans=1220
唔該!!!!!!!!!!!!!!!!!!!!!!!!!:)))
F.2數學難題!!!急!!:$
2012-07-31 1:22 am
回答 (2)
2012-07-31 2:59 am
✔ 最佳答案
Fa1. (x square+3x)square+3(x square+3x)+2Let y=(x square+3x)
so that,
y square+3y square +2
=(y+2)(y+1)
=(x square+3x+2)(x square+3x+1)
=(x+1)(x+2)(x square+3x+1)
Fa2. 2x cube - 3x square +1
=2x cube - 2x -x square +1
=2x square(x-1)-(x square -1)
=2x square(x-1)-(x 1)(x+1)
=(x-1)(2x square-x-1)
=(x-1)(x-1)(2x+1)
Fo1.q=the sum of the positive integers from 11to50
=11+12+13+14+...+50
=(1+2+3+4+...+50)-(1+2+3+4+...+10)
=50(50+1)/2 - 10(10+1)/2
=25x51-5x11
=1220
2012-07-31 22:19:44 補充：
修改:
Fa2. 2x cube - 3x square +1
=2x cube - 2x square -x square +1
=2x square(x-1)-(x square -1)
=2x square(x-1)-(x 1)(x+1)
=(x-1)(2x square-x-1)
=(x-1)(x-1)(2x+1)
參考: me, me
2012-07-31 2:59 am
^ = 次方
1)
Let x^2+3x be y
[x^2+3x)]^2 + 3(x^2+3x) +2
=y^2+3y+2
=(y+2)(y+1)
=(x^2+3x+2)(x^2+3x+1)
=(x+2)(x+1)(x^2+3x+1)
2)
Let f(x)=2x^3-2x^2+1
f(1)=0
(x-1)is a factor of f(x)
f(x)=(x-1)(2x^2+ax-1)
*****0=-a-1
a=-1
f(x)=(x-1)(2x^2-x-1)
=(x-1)(x-1)(2x+1)*****(can be replace by long division)
=(x-1)^2(2x+1)
1)
[(11+50)(40)]/2
=1220
1)
Let x^2+3x be y
[x^2+3x)]^2 + 3(x^2+3x) +2
=y^2+3y+2
=(y+2)(y+1)
=(x^2+3x+2)(x^2+3x+1)
=(x+2)(x+1)(x^2+3x+1)
2)
Let f(x)=2x^3-2x^2+1
f(1)=0
(x-1)is a factor of f(x)
f(x)=(x-1)(2x^2+ax-1)
*****0=-a-1
a=-1
f(x)=(x-1)(2x^2-x-1)
=(x-1)(x-1)(2x+1)*****(can be replace by long division)
=(x-1)^2(2x+1)
1)
[(11+50)(40)]/2
=1220
收錄日期: 2021-05-03 02:55:17
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120730000051KK00584