e.g. x^2 + 2x(a+b) + (a+b)^2
=[x+(a+b)]^2
=(x+a+b)^2 //
問題: x^2 + 4x(y+z) + 4(y+z)^2
請依e.g. 題ge模式作答
我唔知要唔要(y+z) 2次先 定 將個4乘入去先2次.....
大家幫手教一教~ thx =]
容易攞分~factorise
2007-09-07 7:07 am
回答 (5)
2007-09-07 7:17 am
✔ 最佳答案
首先需要明白例題既模式先x^2 + 2x(a+b) + (a+b)^2
= [x+(a+b)]^2
= (x+a+b)^2
如果設x = M , (a+b) =N
M^2 + 2MN + N^2
= (M+N)^2
將M = x , N = (a+b)代番入去
= (x + (a + b)^2
= (x + a + b)^2
將呢個情況改一改
x^2 + 4x(y+z) + 4(y+z)^2
設x = M , (y+z) = N
M^2 + 4MN + (2N)^2
= (M + 2N)^2
將M = x 及 N = (y+z)代番入去
= [x + 2(y+z)]^2
= (x + 2y + 2z)^2 //
2007-09-07 7:15 am
Answer:
x^2 + 4x(y+z)+4(y+z)^2
= x^2 + 2[2x(y+z)] + [2(y+z)]^2
=[x + 2(y+z)]^2
=(x + 2y + 2z)^2 //
即是 2(y+z) = 例題中的 (a+b)
x^2 + 4x(y+z)+4(y+z)^2
= x^2 + 2[2x(y+z)] + [2(y+z)]^2
=[x + 2(y+z)]^2
=(x + 2y + 2z)^2 //
即是 2(y+z) = 例題中的 (a+b)
2007-09-07 7:15 am
by (a+b)^2 = a^2+2ab+b^2
x^2 + 4x(y+z) + 4(y+z)^2
=x^2+2x‧2(y+z)+ [2(y+z)]^2
=[x+2(y+z)]^2
x^2 + 4x(y+z) + 4(y+z)^2
=x^2+2x‧2(y+z)+ [2(y+z)]^2
=[x+2(y+z)]^2
參考: me
2007-09-07 7:12 am
x^2 + 4x(y+z) + 4(y+z)^2
=x^2 +2 (x) [2(z+y)] + [2(y+z)]^2
=[x+ 2(y+z)]^2
=(x+2y+2z)^2
明了吧
2007-09-06 23:15:36 補充：
例如(x a)^2 2n(x a)(y z) n^2(y z)^2=(x a)^2 2(x a)[n(y z)] n^2(y z)^2= [(x a) n(y z)]^2= (x a ny nz)^2
2007-09-06 23:16:59 補充：
無晒ｄ＋ 號 ｔｉｍ慢左小小請多多支持
=x^2 +2 (x) [2(z+y)] + [2(y+z)]^2
=[x+ 2(y+z)]^2
=(x+2y+2z)^2
明了吧
2007-09-06 23:15:36 補充：
例如(x a)^2 2n(x a)(y z) n^2(y z)^2=(x a)^2 2(x a)[n(y z)] n^2(y z)^2= [(x a) n(y z)]^2= (x a ny nz)^2
2007-09-06 23:16:59 補充：
無晒ｄ＋ 號 ｔｉｍ慢左小小請多多支持
2007-09-07 7:10 am
x^2 + 4x(y+z) + 4(y+z)^2
= [x + 2(y+z)]^2
= (x+2y+2z)^2
就係咁啦...唔洗乘左先~~
= [x + 2(y+z)]^2
= (x+2y+2z)^2
就係咁啦...唔洗乘左先~~
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070906000051KK04724