- 8a ^3 q - 16a(aq ^2 - a ^2 q)
_______________________
20a ^2 q ^2
Simplify the numerator and divide.?
2008-07-23 12:39 pm
回答 (6)
2008-07-23 4:11 pm
✔ 最佳答案
[-8a^3q - 16a(aq^2 - a^2q)]/[20a^2q^2]= [-8a^3q - 16a^2q^2 + 16a^3q]/[20a^2q^2]
= [-8a^3q + 16a^3q - 16a^2q^2]/[20a^2q^2]
= 8a^2q[a - 2q]/[20a^2q^2]
= [8a^2q/20a^2q^2][a - 2q]
= [2/5q][a - 2q]
= 2[a - 2q]/5q
= 2/5 * [a - 2q]/q
= 2/5 * [a/q - 2q/q]
= [2/5][a/q - 2]
2008-07-23 7:56 pm
Numerator, N= -8a^3q -16a(aq^2 -a^2q):
Separating out common factors and putting the rest in brackets
- 4a^2q{ 2a + 4(q - a)}
Now
- 4a^2q{ 2a + 4(q - a)}
__________________ =
4a^2q(5q)
[cancel out identical quantities in Num & Denom]
- 2[a + 2(q - a)]
_____________ = (N/D)
5q
Separating out common factors and putting the rest in brackets
- 4a^2q{ 2a + 4(q - a)}
Now
- 4a^2q{ 2a + 4(q - a)}
__________________ =
4a^2q(5q)
[cancel out identical quantities in Num & Denom]
- 2[a + 2(q - a)]
_____________ = (N/D)
5q
2008-07-23 7:54 pm
You forgot laws of indices Panther, haven't you?
Nr = - 8(a^3)q - 16(a^2)(q^2) + 16(a^3)q
= 8(a^3)q - 16(a^2)(q^2)
= 8(a^2)q [ 1 - 2q]
so fraction = 2(1-2q)/5qa^3
I assumed a^3 q to be (a^3)q NOT a^(3q)
Nr = - 8(a^3)q - 16(a^2)(q^2) + 16(a^3)q
= 8(a^3)q - 16(a^2)(q^2)
= 8(a^2)q [ 1 - 2q]
so fraction = 2(1-2q)/5qa^3
I assumed a^3 q to be (a^3)q NOT a^(3q)
2008-07-23 8:14 pm
-8a^3q - 16a(aq^2 - a^2q)
----------------------------------- =
20a^2q^2
-8a^3q - 16a^2q^2 + 16a^3q
----------------------------------------- =
20a^2q^2
8a^3q - 16a^2q^2
----------------------------- =
20a^2q^2
8a^3q - 16a^2q^2
---------- --------------- =
20a^2q^2 20a^2q^2
2a 4
------- - ------- =
5q 5
25q( 2a/5q - 4/5) = 10aq - 20q =
10q(a-2q)
----------------------------------- =
20a^2q^2
-8a^3q - 16a^2q^2 + 16a^3q
----------------------------------------- =
20a^2q^2
8a^3q - 16a^2q^2
----------------------------- =
20a^2q^2
8a^3q - 16a^2q^2
---------- --------------- =
20a^2q^2 20a^2q^2
2a 4
------- - ------- =
5q 5
25q( 2a/5q - 4/5) = 10aq - 20q =
10q(a-2q)
2008-07-23 7:54 pm
= -8a^3 q - 16a^2q^2 + 16a^3q / 20a^2 q^2 (multiply out the bracket)
= 8a^3q - 16 a^2 q^2 / 20a^2 q^2 (simplify)
= 4 a^2q (2a - 4q) / 4a^2 q (5q) (find a common factor on top and bottom)
= 2a- 4q / 5q OR 2(a-2q) / 5q (divide top & bottom by 4 a^2q)
= 8a^3q - 16 a^2 q^2 / 20a^2 q^2 (simplify)
= 4 a^2q (2a - 4q) / 4a^2 q (5q) (find a common factor on top and bottom)
= 2a- 4q / 5q OR 2(a-2q) / 5q (divide top & bottom by 4 a^2q)
2008-07-23 7:53 pm
First: distribute the top:
You get (-8a^3q - 16a^2q^2 + 16a^3q).
Combine like terms and you have 8a^3q - 16a^2q^2.
Factor out 8a^2q.
8a^2q(a-2q)/(20a^2q^2)
Reduce: 2(a-2q)/5q
You get (-8a^3q - 16a^2q^2 + 16a^3q).
Combine like terms and you have 8a^3q - 16a^2q^2.
Factor out 8a^2q.
8a^2q(a-2q)/(20a^2q^2)
Reduce: 2(a-2q)/5q
收錄日期: 2021-05-01 10:52:17
原文連結 [永久失效]:
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