3條f.2升f.3數學(代數分式)

2011-08-16 12:56 am

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吾該晒呀!!! thanks:):):)

回答 (1)

2011-08-16 1:21 am
✔ 最佳答案
1.
[1 / (ab - b²)] - [1 / (a² -ab)]
= [1 / b(a - b)] - [1 / a(a - b)]
= [a / ab(a - b)] - [b / ab(a - b)]
= (a - b) / ab(a - b)
= 1 / ab


2.
[20k / (6j + 5k)] + [24j / (6j + 5k)]
= (20k + 24j) / (6j + 5k)
= (24j + 20k) / (6j + 5k)
= 4(6j + 5k) / (6j + 5k)
= 4


3.
[x² / (x² - y²)] - [y² / (3y² - 3x²)]
= [x² / (x² - y²)] - [y² / 3(y² - x²)]
= [3x² / 3(x² - y²)] + [y² / 3(x² - y²)]
= (3x² + y²) / 3(x² - y²)
[or (3x² + y²) / 3(x + y)(x- y)]
參考: Adam


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