有幾條
(a) ab+bc+2a+2c=
(b) 5xy-10xz+4y-8=
(c) mp二次 +2m+np二次+2n=
Thanks...
數學因式分解有D唔識
2010-01-04 2:08 am
回答 (5)
2010-01-04 2:36 am
✔ 最佳答案
a) ab+bc+2a+2c= b(a+c)+2(a+c)
=(a+c)(b+2)
b) 5xy-10x+4y-8 (唔知你是否打多個z,所以當冇z計)
=5x(y-2)+4(y-2)
=(y-2)(5x+4)
* 如果你冇打錯 :
=5(y-2z)+4(y-2) (冇得計了)
c) mp^2+2m+np^2+2n
=m(p^2+2)+n(p^2+2)
=(p^2+2)(m+n)
2010-01-04 6:06 am
ab+bc+2a+2c
b(a+c)+2(a+c)
(b+2)(a+c)
5xy-10xz+4y-8
5x(y-2z)+4(y-2)
你係唔係打錯??
mp2+2m+np2+2n
(mp2+np2)+(2m+2n)
p2(m+n)+2(m+n)
(p2+2)(m+n)
b(a+c)+2(a+c)
(b+2)(a+c)
5xy-10xz+4y-8
5x(y-2z)+4(y-2)
你係唔係打錯??
mp2+2m+np2+2n
(mp2+np2)+(2m+2n)
p2(m+n)+2(m+n)
(p2+2)(m+n)
2010-01-04 3:04 am
(a) ab+bc+2a+2c
=b(a+c)+2(a+c)
=(a+c)(b+2)
(b)5xy-10xz+4y-8
=5x(y-2z)+4(y-2)
點解多咗個z既?!
如果係咁就不能分解架啦.
如果冇z個答案係:
=(y-2)(5x+4)
(c)mp^2 +2m+np^2+2n
=m (p^2 + 2 ) + n (p^2 + 2 )
=(p^2 + 2 )(m+n)
=b(a+c)+2(a+c)
=(a+c)(b+2)
(b)5xy-10xz+4y-8
=5x(y-2z)+4(y-2)
點解多咗個z既?!
如果係咁就不能分解架啦.
如果冇z個答案係:
=(y-2)(5x+4)
(c)mp^2 +2m+np^2+2n
=m (p^2 + 2 ) + n (p^2 + 2 )
=(p^2 + 2 )(m+n)
2010-01-04 2:46 am
(a.)ab+bc+2a+2c
=b(a+c)+2(a+c)
=(b+2)(a+c)
(b.)5xy-10xz+4y-8
=5x(y-2z)+4(y-2) -------------- 因為 同類項 (y-2z) 和 (y-2) 不相同，所以這個就是答案。
(c.)mp^2+2m+np^2+2n
=p^2(m+n)+2(m+n)
=(p^2+2)(m+n)
希望幫倒你啦！
=b(a+c)+2(a+c)
=(b+2)(a+c)
(b.)5xy-10xz+4y-8
=5x(y-2z)+4(y-2) -------------- 因為 同類項 (y-2z) 和 (y-2) 不相同，所以這個就是答案。
(c.)mp^2+2m+np^2+2n
=p^2(m+n)+2(m+n)
=(p^2+2)(m+n)
希望幫倒你啦！
參考: 我自己
2010-01-04 2:34 am
(a) ab+bc+2a+2c=
A. ab+bc+2a+2c
= b(a+c)+2(a+c)
= (a+c)(b+2)
(b) 5xy-10xz+4y-8=
A. 5xy-10xz+4y-8
= 5x(y-2z)+4(y-2)-----------------Cannot be further factorized. (Problems in question?)
(c) mp二次 +2m+np二次+2n=
A. mp^2+2m+np^2+2n
= m(p^2+2)+n(p^2+2)
= (p^2+2)(m+n)
A. ab+bc+2a+2c
= b(a+c)+2(a+c)
= (a+c)(b+2)
(b) 5xy-10xz+4y-8=
A. 5xy-10xz+4y-8
= 5x(y-2z)+4(y-2)-----------------Cannot be further factorized. (Problems in question?)
(c) mp二次 +2m+np二次+2n=
A. mp^2+2m+np^2+2n
= m(p^2+2)+n(p^2+2)
= (p^2+2)(m+n)
參考: Myself
收錄日期: 2021-04-13 17:02:35
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