Factorize the following expressions:
(0.05x - 2)^2
(4m+0.5n)^2
(a/2 - b/3)^2
(a/2 - 1/a)^2
(-5h^2 - k^2)^2
(2y^2 - 5x^2y)^2
identities (urgent)20pts
2009-08-01 7:38 pm
回答 (2)
2009-08-01 8:08 pm
✔ 最佳答案
(0.05x - 2)^2=0.025x^2 -2^2
=0.0025x^2 -4
(4m+0.5n)^2
=(4m)^2 +(0.5n)^2
=16m^2 +0.25n^2
(a/2 - b/3)^2
=(a/2)^2 -(b/3)^2
=a^2/4 -b^2/9
(a/2 - 1/a)^2
=(a/2)^2 -(1/a)^2
=a^2 /4 -1/a^2
(-5h^2 - k^2)^2
=(-5h^2)^2 -(k^2)^2
=25h^4 -k^4
(2y^2 - 5x^2y)^2
=(2y^2)^2- (5x^2y)^2
=4y^4- 25x^2y+2
參考: myself
2009-08-01 8:12 pm
(0.05x - 2)2
=0.0025x2-0.2x+4
(4m+0.5n)2
=16m2+4mn+0.25n2
(a/2 - b/3)2
=a2/4 - (ab)/3 + b2/9
(a/2 - 1/a)2
=a2/4 - 1 + 1/a2
(-5h2 - k2)2
=25h4+10h2k2+k4
(2y2 - 5x2y)2
=4y4-20x2y3+25x4y2
=0.0025x2-0.2x+4
(4m+0.5n)2
=16m2+4mn+0.25n2
(a/2 - b/3)2
=a2/4 - (ab)/3 + b2/9
(a/2 - 1/a)2
=a2/4 - 1 + 1/a2
(-5h2 - k2)2
=25h4+10h2k2+k4
(2y2 - 5x2y)2
=4y4-20x2y3+25x4y2
收錄日期: 2021-04-23 20:41:46
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090801000051KK00576