利用平方差恆等式/完全平方恆等式展開
1. ( t^2 + 5t )( t+ 5 )
2. ( x-1 )^2 - ( x+1)^2
兩條中二數
2008-10-23 3:28 am
回答 (2)
2008-10-23 7:44 am
✔ 最佳答案
1.(t2 + 5t)(t + 5)
= t(t + 5)(t + 5)
= t(t + 5)2
= t[t2 + 2(t)(5) + (5)2]
= t(t2 + 10t + 25)
= t3 + 10t2 + 25t
=====
2.
方法一：
(x - 1)2 - (x + 1)2
= (x2 - 2x + 1) - (x2 + 2x + 1)
= x2 - 2x + 1 - x2 - 2x - 1
= -4x
方法二：
(x - 1)2 - (x + 1)2
= [(x - 1) - (x + 1)][(x - 1) + (x - 1)]
= [x - 1 - x - 1][x - 1 + x - 1]
= [-2][2x]
= -4x
=
2008-10-24 10:40 pm
1.
( t^2 + 5t )( t+ 5 )
=t^2(t+5)+5t(t+5)
=t^3+5t^2+5t^2+25t
=t^3+10t^2+25t
2.
( x-1 )^2 - ( x+1)^2
=(x^2-2x+1)-(x^2+2x+1)
=x^2-2x+1-x^2-2x-1
=-4x
( t^2 + 5t )( t+ 5 )
=t^2(t+5)+5t(t+5)
=t^3+5t^2+5t^2+25t
=t^3+10t^2+25t
2.
( x-1 )^2 - ( x+1)^2
=(x^2-2x+1)-(x^2+2x+1)
=x^2-2x+1-x^2-2x-1
=-4x
參考: ME
收錄日期: 2021-04-23 23:05:45
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20081022000051KK01494