求下列恆等式中常數A及B的值。
(2A+B)x-3B≡-x+9
展開下列各式。
[ (a/3) + (b/2) ] [ (-a/3) - (b/2) ]
(a-2b+1) (a+2b+1)
Form 2 MATH questions!
2011-05-02 3:15 am
回答 (1)
2011-05-02 3:58 am
✔ 最佳答案
(1) (2A+B)x-3B≡-x+9Put x=0
-3B=9
B=-3
Put x=1
2A+B-3B=8
2A-2B=8
2A+6=8
2A=2
A=1
(2) [ (a/3) + (b/2) ] [ (-a/3) - (b/2) ]
= -[ (a/3) + (b/2) ]^2
= -(a^2/9+ab/3+b^2/4)
= -a^2/9-ab/3-b^2/4
(3) (a-2b+1) (a+2b+1)
=(a+1-2b)(a+1+2b)
=(a+1)^2-4b^2
=a^2+2a+1-4b^2
收錄日期: 2021-04-20 11:54:31
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