factorize the fllowings:(with steps)
1)(p^2)+(1/(p^2)- 2
(ans should be (p-(1/p))^2 )
2)(pq)^2 +(p/q)^2 +(q/p)^2 +(1/pq)^2
(ans should be (p^2 +(1/p^2)) (q^2 + (1/q^2)) )
factorization
2006-12-26 11:44 pm
回答 (2)
2006-12-27 12:54 am
✔ 最佳答案
1. p^2 + [1 / p^2] - 2= (p)^2 - 2(p)(1 / p) + (1 / p)^2
= [p - (1 / p)]^2.......................Here we use a^2 - 2ab + b^2 = (a - b)^2
2. (pq)^2 + (p / q)^2 + (q / p)^2 + (1 / pq)^2
= p^2 q^2 + p^2 / q^2 + q^2 / p^2 + 1 / (p^2 q^2)
= p^2 [q^2 + (1 / q^2)] + (1 / p^2)[q^2 + (1 / q^2)]
= [p^2 + (1 / p^2)][q^2 + (1 / q^2)]
2006-12-27 12:59 am
1) (p^2)+(1/(p^2))- 2
= (p-(1/p))(p-(1/-p)). (ANS)
2) (pq)^2 +(p/q)^2 +(q/p)^2 +(1/pq)^2
= (p^2 +(1/p^2)) (q^2 + (1/q^2)) (ANS)
you should use the cross multiplication method for solving these.
personally, i think that solving these questions doesn't really require steps. instead, you should write out your working on the right hand side.
= (p-(1/p))(p-(1/-p)). (ANS)
2) (pq)^2 +(p/q)^2 +(q/p)^2 +(1/pq)^2
= (p^2 +(1/p^2)) (q^2 + (1/q^2)) (ANS)
you should use the cross multiplication method for solving these.
personally, i think that solving these questions doesn't really require steps. instead, you should write out your working on the right hand side.
收錄日期: 2021-04-13 14:37:46
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061226000051KK02431