1/c 1/d
-------------
c/d - d/c
Simplify the following expression:
2008-08-03 1:27 pm
回答 (8)
2008-08-03 1:31 pm
✔ 最佳答案
assume you mean1/c+ 1/d
-------------
c/d - d/c
= 1/(c-d)
2008-08-03 8:51 pm
so how to solve this problem: (1/c 1/d) / (c/d - d/c)
first find the equivalent of c/d - d/c... put it in one fraction...
= (c/d - d/c)
find the LCD which is dc
= (d - c) / dc
then substitute it in the 1st expression...
= (1/c 1/d) / (c/d - d/c)
= 1/c x 1/d x (dc / d-c)
multiply reciprocal when dividing
cancel cd...
so the final answer is:
1
--------
d - c
hope that helps!!! :)
first find the equivalent of c/d - d/c... put it in one fraction...
= (c/d - d/c)
find the LCD which is dc
= (d - c) / dc
then substitute it in the 1st expression...
= (1/c 1/d) / (c/d - d/c)
= 1/c x 1/d x (dc / d-c)
multiply reciprocal when dividing
cancel cd...
so the final answer is:
1
--------
d - c
hope that helps!!! :)
2008-08-03 8:49 pm
[1/cd]/[(cc-dd)/cd]=1/[cc-dd]
2008-08-03 8:31 pm
0....?
2008-08-03 8:33 pm
1/cd/c^2-d^2/cd
1/c^2-d^2
1/c^2-d^2
2008-08-03 10:33 pm
(1/c + 1/d)/(c/d - d/c)
= (d/cd + c/cd)/(c^2/cd - d^2/cd)
= [(d + c)/cd]/[(c^2 - d^2)/cd]
= [(d + c)/cd][cd/(c^2 - d^2)]
= cd(d + c)/cd(d + c)(c - d) (cancel out cd , d + c)
= 1/(c - d)
= (d/cd + c/cd)/(c^2/cd - d^2/cd)
= [(d + c)/cd]/[(c^2 - d^2)/cd]
= [(d + c)/cd][cd/(c^2 - d^2)]
= cd(d + c)/cd(d + c)(c - d) (cancel out cd , d + c)
= 1/(c - d)
2008-08-03 9:21 pm
1/c + 1/d
-------------
c/d - d/c
multiply top and bottom by cd
(d+c) / (c² - d²)
(d+c) / (c+d)(c-d)
1/(c-d)
.
-------------
c/d - d/c
multiply top and bottom by cd
(d+c) / (c² - d²)
(d+c) / (c+d)(c-d)
1/(c-d)
.
2008-08-03 8:33 pm
1/c + 1/d = (c + d)/cd
Now,
c/d - d/c = (c² - d²)/dc = [(c + d)(c - d)]/dc
Now then:
[(c + d)/cd]/[(c + d)(c - d))/cd]
= [(c + d)/cd][cd/(c + d)(c - d))]
= 1/(c - d)
Hope this helps!
Now,
c/d - d/c = (c² - d²)/dc = [(c + d)(c - d)]/dc
Now then:
[(c + d)/cd]/[(c + d)(c - d))/cd]
= [(c + d)/cd][cd/(c + d)(c - d))]
= 1/(c - d)
Hope this helps!
收錄日期: 2021-05-01 10:56:28
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080803052700AA0VO6l