試展開下列各式:(唔該用恆等式)
A.(a-1/a)的2次
B.(a/3+b/2)(-a/3-b/2)
C.(a-2b+1)(a+2b+1)
s.2 恆等式
2007-11-04 5:34 am
回答 (3)
2007-11-04 5:41 am
✔ 最佳答案
a) (a-1.a)^2 = a^2 - 2(a)(1/a) + 1/a^2= a^2 - 2 + (1/a^2)
b) (a/3+b/2)(-a/3-b/2 )
= -(a/3+b/2)(a/3+b/2 )
=-[(a/3)^2 + 2(a/3)(b/2) + (b/2)^2]
=-[a^2/9 + ab/3 + b^2/4]
=-a^2/9 - ab/3 - b^2/4
c) (a-2b+1)(a+2b+1)
=[(a+1)-2b][(a+1)+2b]
=(a+1)^2 - (2b)^2
= a^2 + 2a+1 - 4b^2
2007-11-04 10:49 pm
A. (a-1/a)^2
=(a-1)^2/a^2 ^2=2次方
=(a^2-2a+1)/a^2
B. (a/3+b/2)(-a/3-b/2)
= -(a/3+b/2)(a/3+b/2)
= -(a^2/9+b^2/4)
C. (a-2b+1)(a+2b+1)
=a是否欠了2次方，否則不能用恆等式，如a沒有2次方，答案是=a^2+2a-4b^2+1
2007-11-04 14:53:29 補充：
如是欠2次方，答案是a^4 2a^2-4b^2 1
2007-11-04 14:54:26 補充：
應該是a^4 2a^2-4b^2 1
=(a-1)^2/a^2 ^2=2次方
=(a^2-2a+1)/a^2
B. (a/3+b/2)(-a/3-b/2)
= -(a/3+b/2)(a/3+b/2)
= -(a^2/9+b^2/4)
C. (a-2b+1)(a+2b+1)
=a是否欠了2次方，否則不能用恆等式，如a沒有2次方，答案是=a^2+2a-4b^2+1
2007-11-04 14:53:29 補充：
如是欠2次方，答案是a^4 2a^2-4b^2 1
2007-11-04 14:54:26 補充：
應該是a^4 2a^2-4b^2 1
參考: myself
2007-11-04 5:43 am
A. (a-1/a)^2
=(a-1)^2/a^2
=a^2-2a+1/a^2
B.(a/3+b/2)(-a/3-b/2)
=-a^2/9-ab/3-b^2/4
C.(a-2b+1)(a+2b+1)
=a^2+2ab+a-2ab-4b^2-2b+a+2b+1
=a^2+2a+1
=(a-1)^2/a^2
=a^2-2a+1/a^2
B.(a/3+b/2)(-a/3-b/2)
=-a^2/9-ab/3-b^2/4
C.(a-2b+1)(a+2b+1)
=a^2+2ab+a-2ab-4b^2-2b+a+2b+1
=a^2+2a+1
收錄日期: 2021-04-13 18:12:24
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