help! f.2 maths [復制鏈接]

2ab-a 二次 -b 二次 +9=?

i have the answer
but a want列式
thz

2ab-a*a-b*b+9
=-(a*a-2ab+b*b)+9
=-(a-b)二次+9

下面是引用kch於2006-01-06 20:22發表的:
2ab-a*a-b*b+9
=-(a*a-2ab+b*b)+9
=-(a-b)二次+9

but my answer is (3-a+b)(3+a-b).......

2ab-a^2-b^2+9
=-(a^2-2ab+b^2)+9
=-(a-b)^2+9
=-[(a-b)^2-3^2]
=-[(a-b+3)(a-b-3)
=(3+a-b)(3-a+b)

下面是引用啊謝於2006-01-06 20:44發表的:
2ab-a^2-b^2+9
=-(a^2-2ab+b^2)+9
=-(a-b)^2+9
=-[(a-b)^2-3^2]
=-[(a-b+3)(a-b-3)
.......

ys
this is la
thz

下面是引用Tam Chun O於2006-01-06 20:54發表的:
ys
this is la
thz

o弟錯左...
sorry

下面是引用啊謝於2006-01-06 20:44發表的:
2ab-a^2-b^2+9
=-(a^2-2ab+b^2)+9
=-(a-b)^2+9
=-[(a-b)^2-3^2]
=-[(a-b+3)(a-b-3)
.......

i don't know this
=-[(a-b+3)(a-b-3)]
=(3+a-b)(3-a+b)
thz

help ar

i am exam la
help ar!!!!!!!

下面是引用Tam Chun O於2006-01-06 21:49發表的:
i don't know this
=-[(a-b+3)(a-b-3)]
=(3+a-b)(3-a+b)
thz

=-[(a-b+3)(a-b-3)]
=[-a-(-b)-(-3)](3+a-b)
=(3-a+b)(3+a-b)

今次一定無錯！！！

下面是引用kch於2006-01-06 22:29發表的:
=-[(a-b+3)(a-b-3)]
=[-a-(-b)-(-3)](3+a-b)
=(3-a+b)(3+a-b)

why +9 have to change to -9?

下面是引用Tam Chun O於2006-01-06 22:31發表的:
why +9 have to change to -9?

邊度有－９？

下面是引用kch於2006-01-06 22:34發表的:
邊度有－９？

=-[(a-b+3)(a-b-3)]

+3x-3 =-9

kch
can you 列式 again?
thz

下面是引用Tam Chun O於2006-01-06 22:36發表的:
=-[(a-b+3)(a-b-3)]
+3x-3 =-9

ｎｏ
-[(a-b+3)(a-b-3)]
將個負號放落(a-b+3)度
就變左-a+b-3
咁(-3)*(-3)= 9