~~~2
(y-1) -------------- (y-1) 的 square
= (y-1)(y-1)
= ??
(y-1)(y-1)=?
2007-07-09 11:57 pm
回答 (7)
2007-07-11 1:38 am
✔ 最佳答案
(y-1)(y-1)=y^2-y-y+1
=y^2-2y+1
If y=3,
(3-1)*(3-1)
=2*2
=4
or
=3^2 -6+1
=9-6+1
=4
lfy=5
(5*1)(5*1)
=4*4
=16
or
=5^2-10+1
=25-10+1
=16
唔信嘅就試吓啦!
HOPE CAN HELP YOU!
參考: ME
2007-07-11 4:04 am
(y - 1) (y - 1)
= (y - 1)^2
= y^2 - 2 (y) (1) + 1
= y^2 - 2y + 1
= (y - 1)^2
= y^2 - 2 (y) (1) + 1
= y^2 - 2y + 1
2007-07-10 3:53 am
(y-1)(y-1)
=(y-1)^2
or
(y-1)(y-1)
=y(y-1)-1(y-1)
=y^2-y-y+2
=y^2-2y+2
if find y, there is no solutions
=(y-1)^2
or
(y-1)(y-1)
=y(y-1)-1(y-1)
=y^2-y-y+2
=y^2-2y+2
if find y, there is no solutions
2007-07-10 1:56 am
利用恆等式,答案係y^2-2y+2
詳細d就係咁
(y-1)(y-1)
=(y-1)(y)+(y-1)(-1)
=y^2-y-y+2
=y^2-2y+2
2007-07-11 17:58:02 補充：
唔好意思,答案應該係y^2-2y 1
詳細d就係咁
(y-1)(y-1)
=(y-1)(y)+(y-1)(-1)
=y^2-y-y+2
=y^2-2y+2
2007-07-11 17:58:02 補充：
唔好意思,答案應該係y^2-2y 1
收錄日期: 2021-04-12 21:47:44
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https://hk.answers.yahoo.com/question/index?qid=20070709000051KK02686