(-a^2b/b^2c) divided by (a^2c/B^2c^2)?
回答 (2)
2008-07-09 4:36 pm
✔ 最佳答案
Dividing fractions is the same as muyltiplying by the reciprocal, so:(-a²b / b²c) / (a²c / b²c²)
(-a²b / b²c) * (b²c² / a²c)
Now multiply the fractions:
-a²b³c² / a²b²c²
Now simply:
-b
2008-07-09 4:36 pm
(-a^2b/b^2c)/(a^2c/b^2c^2)
= (-a^2/bc)/(a^2/b^2c)
= [-a^2/bc][1/(a^2/b^2c)]
= [-a^2/bc][1(b^2c/a^2)]
= [-a^2/bc][b^2c/a^2]
= -a^2b^2c/a^2bc
= (-a^2/a^2)(b^2/b)(c/c)
= -1(b)(1)
= -b
= (-a^2/bc)/(a^2/b^2c)
= [-a^2/bc][1/(a^2/b^2c)]
= [-a^2/bc][1(b^2c/a^2)]
= [-a^2/bc][b^2c/a^2]
= -a^2b^2c/a^2bc
= (-a^2/a^2)(b^2/b)(c/c)
= -1(b)(1)
= -b
收錄日期: 2021-05-01 10:47:46
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080709082749AAfG2nb