x(x+1)分之ㄧ+(x+1)(x+3)分之二+(x+3)(x+6)分之三=x(x+6)分之k,試求常數k的值?
回答 (2)
2016-06-15 3:12 pm
1/(x(x+1)) + 2/((x+1)(x+3)) + 3/((x+3)(x+6))
= (x+1 - x)/(x(x+1)) + (x+3 - (x+1))/((x+1)(x+3)) + (x+6 - (x+3))/((x+3)(x+6))
= 1/x - 1/(x+1) + 1/(x+1) - 1/(x+3) + 1/(x+3) - 1/(x+6)
= 1/x - 1/(x+6)
= 6/(x(x+6))
常數 k = 6
= (x+1 - x)/(x(x+1)) + (x+3 - (x+1))/((x+1)(x+3)) + (x+6 - (x+3))/((x+3)(x+6))
= 1/x - 1/(x+1) + 1/(x+1) - 1/(x+3) + 1/(x+3) - 1/(x+6)
= 1/x - 1/(x+6)
= 6/(x(x+6))
常數 k = 6
2016-06-28 4:00 am
1/[x(x+1)]+2/[(x+1)(x+3)]+3/[(x+3)(x+6)=k/[x(x+6)]
left=[(x+3)(x+6)+2x(x+6)+3x(x+1)]/[x(x+1)(x+3)(x+6)
left=(x^2+9x+18+2x^2+12x+3x^3+3x)/[x(x+1)(x+3)(x+6)]
left=(6x^2+24x+18)/[x(x+1)(x+3)(x+6)]
left=6(x^2+4x+3)/[x(x+1)(x+3)(x+6)]
left=6(x+1)(x+3)/[x(x+1)(x+3)(x+6)]
left=6/[x(x+6)
k=6
left=[(x+3)(x+6)+2x(x+6)+3x(x+1)]/[x(x+1)(x+3)(x+6)
left=(x^2+9x+18+2x^2+12x+3x^3+3x)/[x(x+1)(x+3)(x+6)]
left=(6x^2+24x+18)/[x(x+1)(x+3)(x+6)]
left=6(x^2+4x+3)/[x(x+1)(x+3)(x+6)]
left=6(x+1)(x+3)/[x(x+1)(x+3)(x+6)]
left=6/[x(x+6)
k=6
收錄日期: 2021-04-18 15:11:53
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160615065612AA4RyOf