展開下列各式
1. 12(x/4+y/2)(x/4-y/2)
2. (1/2-u^2/4)(u^2/4+1/2)
3. -(t^2/3+4)(4-t^2/3)
4. b(4b-a/2)(4b+a/2)
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2012-08-03 8:15 pm
回答 (2)
2012-08-03 8:42 pm
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2012-08-03 12:45:03 補充：
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參考: My Maths World
2012-08-04 6:47 am
1.
12(x/4+y/2)(x/4-y/2)
=12[(x^2)/16][(y^2)/4]
=(3x^2)/4-3y^2
2.
(1/2-u^2/4)(u^2/4+1/2)
=1/4-u^4/16
3.
-(t^2/3+4)(4-t^2/3)
=-(16-t^4/9)
=t^4/9-16
4.
b(4b-a/2)(4b+a/2)
=b(16b^2-a^2/4)
=16b^3-a^2/4
12(x/4+y/2)(x/4-y/2)
=12[(x^2)/16][(y^2)/4]
=(3x^2)/4-3y^2
2.
(1/2-u^2/4)(u^2/4+1/2)
=1/4-u^4/16
3.
-(t^2/3+4)(4-t^2/3)
=-(16-t^4/9)
=t^4/9-16
4.
b(4b-a/2)(4b+a/2)
=b(16b^2-a^2/4)
=16b^3-a^2/4
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原文連結 [永久失效]:
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