(6-sqrt(5))/(6+sqrt(5))
Is the answer 31/41? If not where did I go wrong?
Rationalise the denominator?
2018-04-21 2:42 am
回答 (7)
2018-04-21 7:30 am
(6 - sqrt(5)) / (6 + sqrt(5))
= (6 - sqrt(5))^2 / (6 + sqrt(5)) (6 - sqrt(5))
= (41 - 12 sqrt(5)) / 31
The answer is not 31/41.
= (6 - sqrt(5))^2 / (6 + sqrt(5)) (6 - sqrt(5))
= (41 - 12 sqrt(5)) / 31
The answer is not 31/41.
2018-04-21 7:04 am
(6-√5)/(6+√5) = a + b√5
6-√5 = (6+√5)(a+b√5) = (6a+5b) + (a+6b)√5
6a+5b=6 and a+6b=-1, so a = 41/31 and b = -12/31
(6-√5)/(6+√5) = (41 - 12√5)/31
6-√5 = (6+√5)(a+b√5) = (6a+5b) + (a+6b)√5
6a+5b=6 and a+6b=-1, so a = 41/31 and b = -12/31
(6-√5)/(6+√5) = (41 - 12√5)/31
2018-04-21 4:48 am
(6-√5)/(6+√5)
multiply and divide by (6-√5)
= (6-√5)(6-√5) /((6+√5)(6-√5))
The denominator is of the form (a+b)(a-b)=a^2-b^2
a=6
a^2=36
b=√5
b^2=5
a^2-b^2 =36-5 = 31
= (6-√5)(6-√5) / 31
= (36-6√5-6√5+5) /31
= (36-12√5+5) /31
= (41-12√5)/31
multiply and divide by (6-√5)
= (6-√5)(6-√5) /((6+√5)(6-√5))
The denominator is of the form (a+b)(a-b)=a^2-b^2
a=6
a^2=36
b=√5
b^2=5
a^2-b^2 =36-5 = 31
= (6-√5)(6-√5) / 31
= (36-6√5-6√5+5) /31
= (36-12√5+5) /31
= (41-12√5)/31
2018-04-21 4:23 am
(6 - sqrt(5)) / (6 + sqrt(5))
= (6 - sqrt(5))^2 / (6 + sqrt(5))(6 - sqrt(5))
= (36 - 12 sqrt(5) + 5) / (36 - 5)
= (41 - 12 sqrt(5)) / 31. Final.
= (6 - sqrt(5))^2 / (6 + sqrt(5))(6 - sqrt(5))
= (36 - 12 sqrt(5) + 5) / (36 - 5)
= (41 - 12 sqrt(5)) / 31. Final.
2018-04-21 2:53 am
(6-sqrt(5))/(6+sqrt(5) = (6-sqrt(5))^2/(36 - 5) = (36 -12sqrt(5) + 5)/31 =
(41-12sqrt(5))/31
(41-12sqrt(5))/31
2018-04-21 2:45 am
Multiply both the numerator and denominator by [6 - sqrt(5)].
You get
(6 - sqrt(5))^2 / (36 - 5)
= [36 + 5 - 12*sqrt(5)] / 31
= (41 - 12*sqrt(5))/31.
You get
(6 - sqrt(5))^2 / (36 - 5)
= [36 + 5 - 12*sqrt(5)] / 31
= (41 - 12*sqrt(5))/31.
收錄日期: 2021-04-24 01:00:32
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20180420184227AAb63IK