化簡代數分式!
(1/a+b + 1/a-b) x (1 - b/a)
(1 + 1/a-2) / (1 - a/a+1)
thx!
F.2 MATHS!
2006-11-15 7:32 am
回答 (4)
2006-11-15 7:42 am
✔ 最佳答案
(1/a+b + 1/a-b) x (1 - b/a)=[(a-b+a+b)/(a+b)(a-b)][(a-b)/a]
=2a/a(a+b)
=2/(a+b)
(1 + 1/a-2) / (1 - a/a+1)
=[(a-2+1)/(a-2)]/[(a+1-a)/(a+1)]
=[(a-1)/(a-2)]/[1/(a+1)]
=(a-1)(a+1)/(a-2)
2006-11-15 7:43 am
(1/a+b + 1/a-b) x (1 - b/a)
= [(a-b)+(a+b)/(a+b)(a-b)][(a-b)/a]
= [2a/(a+b)(a-b)][(a-b)/a]
= 2/(a+b)
(1 + 1/a-2) / (1 - a/a+1)
={[ (a-2)+1]/(a-2)}/{[(a+1)-a]/(a+1)}
= [(a-1)/(a-2)]/[1/(a+1)]
= (a-1)(a+1)/(a-2)
= [(a-b)+(a+b)/(a+b)(a-b)][(a-b)/a]
= [2a/(a+b)(a-b)][(a-b)/a]
= 2/(a+b)
(1 + 1/a-2) / (1 - a/a+1)
={[ (a-2)+1]/(a-2)}/{[(a+1)-a]/(a+1)}
= [(a-1)/(a-2)]/[1/(a+1)]
= (a-1)(a+1)/(a-2)
2006-11-15 7:43 am
(1/a+b + 1/a-b) x (1 - b/a)
= [a-b+a+b / (a+b)(a-b)] x (a-b/a)
= 2a / (a+b)(a-b) x (a-b/a)
= 2a(a-b) / a(a+b)(a-b)
= 2 / (a+b)
(1 + 1/a-2) / (1 - a/a+1)
= (a-2+1 / a-2) / (a+1-a/a+1)
= (a-1 / a-2)(1 / a+1)
= (a-1) / (a-2)(a+1)
= [a-b+a+b / (a+b)(a-b)] x (a-b/a)
= 2a / (a+b)(a-b) x (a-b/a)
= 2a(a-b) / a(a+b)(a-b)
= 2 / (a+b)
(1 + 1/a-2) / (1 - a/a+1)
= (a-2+1 / a-2) / (a+1-a/a+1)
= (a-1 / a-2)(1 / a+1)
= (a-1) / (a-2)(a+1)
2006-11-15 7:40 am
=(1/a + b + 1/a -b) X 1- b/a
={(1/a + b/1 + 1/a - b/1 ) X 1}- b/a
=2/a - b/a
=(2-b)/a
=(1+ 1/2 -2 ) /(1-a/a +1)
=(3/2-4/2 ) /(2-1)
= - 1/2
is it correct?
={(1/a + b/1 + 1/a - b/1 ) X 1}- b/a
=2/a - b/a
=(2-b)/a
=(1+ 1/2 -2 ) /(1-a/a +1)
=(3/2-4/2 ) /(2-1)
= - 1/2
is it correct?
收錄日期: 2021-04-12 20:50:52
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061114000051KK05290