which of the following(s) is/are true?
1.(-ab+mn)^2=(ab-mn)^2
2.(-ab-mn)^2=-(ab+mn)^2
the answer should be 1.But I don't know why.Please show me step.
___1____ + ___2x_____
y-2x 4x^2-y^2
the answer should be - ____y_____
4x^2-y^2
but I don;t know why I get another answer
____y_____
4x^2-y^2
Please show me step
maths correction
2007-01-19 2:20 am
回答 (2)
2007-01-19 2:30 am
✔ 最佳答案
1 is true, 2 is falseLHS=(-ab+mn)2
=(-1)2(ab-mn)2
=(ab-mn)2
=RHS
but 2):
LHS=(-ab-mn)2
=(-1)2(ab+mn)2
=(ab+mn)2
≠RHS (=-(ab+mn)2)
Your second qusetion:
1/(y-2x) + 2x/(4x2-y2)
=1/(y-2x) + 2x/[(2x)2-y2]
=1/(y-2x) + 2x/(2x+y)(2x-y)
=1/(y-2x) - 2x/(2x+y)(y-2x)
=(2x+y)/(2x+y)(y-2x) - 2x/(2x+y)(y-2x)
=(2x+y-2x)/(2x+y)(y-2x)
=y/(y+2x)(y-2x)
=y/[y2-(2x)2]
=y/(y2-4x2)
(or =-y/(4x2-y2) )
參考: me
2007-01-19 2:38 am
左方:(-ab+mn)^2 *1
=(-ab)^2+2(-ab)(mn)+(mn)^2 *2
=a^2 b^2 - 2abmn + m^ 2n^2
右方:(ab-mn)^2
=(ab)^2 - 2(ab)(mn) + (mn)^2
=a^2 b^2 - 2abmn+ m^2 n^2
左方=右方
所以 (-ab+mn)^2=(ab-mn)^2是恆等式
注意********** 1 因式分解
2 負負得正 (-ab)^2 = a^2b^2
2007-01-18 18:42:15 補充：
左方:(-ab mn)^2旁的星唔係x 1 x2係備注
=(-ab)^2+2(-ab)(mn)+(mn)^2 *2
=a^2 b^2 - 2abmn + m^ 2n^2
右方:(ab-mn)^2
=(ab)^2 - 2(ab)(mn) + (mn)^2
=a^2 b^2 - 2abmn+ m^2 n^2
左方=右方
所以 (-ab+mn)^2=(ab-mn)^2是恆等式
注意********** 1 因式分解
2 負負得正 (-ab)^2 = a^2b^2
2007-01-18 18:42:15 補充：
左方:(-ab mn)^2旁的星唔係x 1 x2係備注
收錄日期: 2021-04-12 21:32:52
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