請問~中一Maths_代數式

2009-10-22 6:29 am

代數式~~~~~~~~~

(4a+2)-(a)b

P.S:(a)是二次方

假設a=1,b=3

問答案是...

請告訴我計算的過程...please~~~~~~~

更新1:

做錯題目了... 是a有二次方...不是b...

回答 (6)

2009-10-22 7:23 am

✔ 最佳答案

(4a^2+2)-(a^2)b
Sub. a=1 and b=3,
[4(1^2)+2)-(1^2)(3)
=(4x1+2)-(1)(3)
=4+2-3
=3

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2009-10-23 6:01 am

假設a=1,b=3
(4a+2)-(a)b
=[4(1)+2]-(1)(3)
=6-3
=3

參考: myself

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2009-10-23 3:25 am

(4a+2)-(a^2)b
=(4x1+2)-(1^2)3
=6-1x3
=6-3
=3

參考: 我個腦(可能有錯咖)

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2009-10-23 3:20 am

(4a+2)-(a)b
a=1,b=3(given)

(4(1)+2)-(1^2)(3)
=6-3
=3

參考: me

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2009-10-22 7:23 am

(4a+2)-a^2*b
= (4*1+2)-(1^2)*3
= 6-3
=3

參考: Myself

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2009-10-22 6:46 am

是(4a+2)-b^2嗎?

(4a+2)-b^2
=(4x1+2)-3^2
=(4+2)-9
=6-9
=-3

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