1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)=A/x(x+3)
求A=?
正解為3
雙根號化簡
大根號8-4小根號3=?
正解 根號6-根號2
急!高一數學!部分分式法和雙重根號的化簡!
2014-05-03 6:11 pm
回答 (13)
2014-05-03 8:09 pm
✔ 最佳答案
1/[x(x+1)]+1/[(x+1)(x+2)]+1/[(x+2)(x+3)]=A/[x(x+3)]，求A=?Sol
1/[x(x+1)]+1/[(x+1)(x+2)]+1/[(x+2)(x+3)]
=[(x+1)－x]/[x(x+1)]+[(x+2)－(x+1)]/[(x+1)(x+2)]+[(x+3)－(x+2]/[(x+2)(x+3)]
=[1/x－1/(x+1)]+[1/(x+1)－1/(x+2)]+[1/(x+2)－1/(x+3]
=1/x－1/(x+3)
=[(x+3)－x]/[x(x+3)]
=3/[x(x+3)]
A=3
雙根號化簡
√(8－4√3)?
Sol
√(8－4√3)
=√(8－2√12)
=√(6－2√12+2)
=√[(√6)^2－2√6*√2+(√2)^2]
=√[(√6－√2)^2]
=√6－√2
2014-06-21 8:23 am
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2014-05-29 3:13 pm
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2014-05-03 6:49 pm
(1)
X=1, 1/2 + 1/6 + 1/12 = A/4 A = 3
(2)
√((a+b)±2√ab) = √a±√b
√(8-4√3) = √(8-2√12) = √((6+2)-2√6x2) = √6 - √2
X=1, 1/2 + 1/6 + 1/12 = A/4 A = 3
(2)
√((a+b)±2√ab) = √a±√b
√(8-4√3) = √(8-2√12) = √((6+2)-2√6x2) = √6 - √2
收錄日期: 2021-04-30 18:42:53
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140503000016KK02666