我想問 (a-3)(a-1) 等如幾多?
我個朋友問我,但我又唔識計,請大家幫幫手
數學問題...
2006-11-15 3:45 am
回答 (15)
2006-11-15 3:48 am
✔ 最佳答案
a^-4a+3(a-3)(a-1)
= a*a - a*1 - 3*a + (-3)*(-1)
= a的二次方 -4a +3
2006-11-15 4:40 am
(a-3)(a-1)
=a^2-a-3a+3
=a^2-4a+3
=a^2-a-3a+3
=a^2-4a+3
2006-11-15 4:32 am
(a-3)(a-1)
=a^ -1a-3a+1
=a^ -1a-3a+1
2006-11-15 4:22 am
(a-3)(a-1) = a^2 - 3a - a + 3 = a^2 - 4a +3
* a ^ 2 = a 的二次方.
Tis method is called expansion.It means we need to make the factor (the 括號) to be the simpliest answer.
In order to do that, you should know the format:
( A + B) ( C + D) = AC + BC + AD + BD
Then , use the method to think about your question.
Remember that whenever there're same monomial (同項) , simplify them!!!!!!!
* a ^ 2 = a 的二次方.
Tis method is called expansion.It means we need to make the factor (the 括號) to be the simpliest answer.
In order to do that, you should know the format:
( A + B) ( C + D) = AC + BC + AD + BD
Then , use the method to think about your question.
Remember that whenever there're same monomial (同項) , simplify them!!!!!!!
2006-11-15 4:14 am
(a-3)(a-1)
= a2– a – 3a + 3
= a2 –4a+3
* that a2 = a xa , i don't know how to type two square
= a2– a – 3a + 3
= a2 –4a+3
* that a2 = a xa , i don't know how to type two square
2006-11-15 3:57 am
(a-3)(a-1)
=a^2-4a+3
因為(x-y)(z-a)
=xz-ax-yz+ay
所以(a-3)(a-1)
=a^2-a-3a+3
=a^2-4a+3
=a^2-4a+3
因為(x-y)(z-a)
=xz-ax-yz+ay
所以(a-3)(a-1)
=a^2-a-3a+3
=a^2-4a+3
2006-11-15 3:51 am
（a-3)(a-1)
=a２次－３a-a+3
=a２次-4a+3
=a２次－３a-a+3
=a２次-4a+3
2006-11-15 3:49 am
a=什麼？
2006-11-15 3:49 am
(a-3)(a-1)
a-3=0 or a-1=0
a=3 or a=1
a-3=0 or a-1=0
a=3 or a=1
2006-11-15 3:49 am
(a-3)(a-1)
=a(a-1)-3(a-1)
=a^2-a-3a+3
=a^2-4a+3
^ stand for "to the power "..
=a(a-1)-3(a-1)
=a^2-a-3a+3
=a^2-4a+3
^ stand for "to the power "..
收錄日期: 2021-04-12 22:51:36
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