(y-1)(y-1)=?

(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!

(y - 1) (y - 1)
= (y - 1)^2
= y^2 - 2 (y) (1) + 1
= y^2 - 2y + 1

(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

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

(y-1)^2
=y^2-2(y)(1)+1^2
=y^2-2y+1

=y^2-y-y+1

=y^2-2y+1

=y square-y-y+1
=y square-2y+1

so easy??
有唔明再問....