中二數學一問(4)

2007-12-07 1:59 am
可以用因式分解或併項計算
2a.xy+xz+6y+6z
2c.3xy-3xz+5y-5z
2e.2am-2an-7bm+7bn
3a.6ac-bd-6ad+bc
3c.2a二次-9abc-6ab+3ac
4a.r二次-rs-rt+st
4c.6g-g二次+6h-9h
5a.y三次+y二次+y=1
5c.2+3a-14a二次-21a三次
更新1:

5a.y三次+y二次+y+1

回答 (3)

2007-12-07 2:27 am
✔ 最佳答案
2a. xy+xz+6y+6z
=(xy+xz)+(6y+6z)
=x(y+z)+6(y+z)
=(y+z)(x+6)


2c. 3xy-3xz+5y-5z
=(3xy-3xz)+(5y-5z)
=3x(y-z)+5(y-z)
=(y-z)(3x+5)

2e. 2am-2an-7bm+7bn
=(2am-2an)-(7bm+7bn)
=2a(m-n)-7b(m-n)
=(m-n)(2a-7b)

3a.6ac-bd-6ad+bc
=(6ac-6ad)-(bd+bc)
=6a(c-d)-b(d-c)
=6a(c-d)-b(c-d)
=(c-d)(6a-b)

3c. 2a^2-9abc-6ab+3ac
=(2a^2-6ab)-(9abc+3ac)
=2a(a-3b)-3ac(3b+a)
=(a-3b)(2a-3ac)

4a. r^2-rs-rt+st
=(r^2-rs)-(rt+st)
=r(r-s)-t(r-s)
=(r-s)(r-t)

4c. 6g-g^2+6h-9h
=6g-g^2-3h

5a. y^3+y^2+y=1
>>(y^2+1)(y-1)=0
y-1+0 or y^2+1=0
y+1 or y^= -1

5c. 2+3a-14a^2-21a^3
=2(1-7a^2)+3a(1-7a^2)
=(1-7a^2)(2+3a)



終於做完了,
很辛苦的,
每個step 做 gar!
P.S. ^ = to the power
希望幫到你na=]
參考: MYSELF
2007-12-07 2:09 am
2a. xy+xz+6y+6z
=x(y+z)+6(y+z)
=(x+6)(y+z)
2c.3xy-3xz+5y-5z
=3x(y-z)+5(y-z)
=(3x+5)(y-z)

其餘的都差不多
參考: 我自己
2007-12-07 2:03 am
2a.xy+xz+6y+6z


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