F. 2 Maths-Questions about factorization by using identities (最佳解答取40點)
Factorize the following expressions.
1. b^2+ab+1/4a^2
2. 3+9/a^2+a^2/4
Please show your working steps clearly and explain them briefly.
Tell me about the techniques to deal with these problems too if you can.
F. 2 Maths-Factorization by using identities(兩題=+40點)
2006-12-20 6:26 am
回答 (4)
2006-12-20 6:38 am
✔ 最佳答案
1)b^2+ab+1/4a^2=^2+2(a/2)(b)+(a/2)^2
=b-0.5a)^2 其實一眼睇到=.="
2)3+9/a^2+a^2/4
(3/a)^2+2(3/a)+(a/2)^2
=(3/a+a/2)^2
2006-12-19 22:40:45 補充：
有d字唔見左1)b^2 ab 1/4a^2=b^2 2(a/2)(b) (a/2)^2=(b-0.5a)^2 其實一眼睇到=.="2)3 9/a^2 a^2/4(3/a)^2 2(3/a) (a/2)^2=(3/a a/2)^2都係一眼就睇到=.="其實d step係睇到之後再想出來
2006-12-20 9:22 am
1. b^2+ab+1/4a^2 <-----將1/4a^2 變做2次方的format
=b^2+ab+(a/2)^2<------用恆等式計
=b^2+ 2(b)(a/2) + (a/2)^2
=(b+1/2a)^2
2.3+9/a^2+a^2/4
=9/a^2+3+a^2/4<------排返正確2次方的formay先
=(3/a)^2+3+(a/2)^2<--------抽2次方
=(3/a)^2 + 2(3/a)(a/2) + (a/2)^2 <----------用恆等式計
=(3/a +a/2)^2
希望幫到你..=)
=b^2+ab+(a/2)^2<------用恆等式計
=b^2+ 2(b)(a/2) + (a/2)^2
=(b+1/2a)^2
2.3+9/a^2+a^2/4
=9/a^2+3+a^2/4<------排返正確2次方的formay先
=(3/a)^2+3+(a/2)^2<--------抽2次方
=(3/a)^2 + 2(3/a)(a/2) + (a/2)^2 <----------用恆等式計
=(3/a +a/2)^2
希望幫到你..=)
2006-12-20 6:42 am
1) (b+1/2a)(b+1/2a)
=(b+1/2a)^2
2) 3+36/4a^2+a^4/4a^2=3+9/a^2+a^2/4
(36+a^4)/4a^2+3=3+9/a^2+a^2/4
(36+a^4)/4a^2=9/a^2+a^2/4
(36+a^4)=
2006-12-19 22:46:37 補充：
sor.......第2題因為冇紙所以好難睇黎計= ="而且我唔知係factorization.......上面果個計得岩....!~
=(b+1/2a)^2
2) 3+36/4a^2+a^4/4a^2=3+9/a^2+a^2/4
(36+a^4)/4a^2+3=3+9/a^2+a^2/4
(36+a^4)/4a^2=9/a^2+a^2/4
(36+a^4)=
2006-12-19 22:46:37 補充：
sor.......第2題因為冇紙所以好難睇黎計= ="而且我唔知係factorization.......上面果個計得岩....!~
參考: 自己= =
2006-12-20 6:41 am
1. b^2+ab+1/4a^2 <-----打1/4a^2 抽個2次方出哩
=b^2+ab+(a/2)^2<------用恆等式計
=(b+a/2)^2
2.3+9/a^2+a^2/4
=9/a^2+3+a^2/4<------排返正確位置
=(3/a)^2+3+(a/2)^2<--------抽2次方,並用恆等式
=(3/a +a/2)^2
=b^2+ab+(a/2)^2<------用恆等式計
=(b+a/2)^2
2.3+9/a^2+a^2/4
=9/a^2+3+a^2/4<------排返正確位置
=(3/a)^2+3+(a/2)^2<--------抽2次方,並用恆等式
=(3/a +a/2)^2
收錄日期: 2021-04-12 22:35:49
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