(2m+n-p)(2m-n+p)==?
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(2m+n-p)(2m-n+p)==?(2m+n-p)(2m-n+p)==?(2m+n-p)(2m-n+p)==?原式=[2m+(n-p)][2m-(n-p)]=(2m)²-(n-p)&su
(2m+n-p)(2m-n+p)==?
(2m+n-p)(2m-n+p)==?
(2m+n-p)(2m-n+p)==?
原式=[2m+(n-p)][2m-(n-p)]
=(2m)²-(n-p)²
=4m²-(n²-2pn+p²)
=4m²-n²+2pn-p²
=[2m+(n-p)][2m-(n-p)]
=4m^2-(n-p)^2
m-(2m-n-p)=
若 P=2m,m=2n,则m+p-n等于()
(2m+n-p)( 2m+p-n)
已知有理数m,n,p,q在数轴上的位置如图所示,且|m|=|n| ,化简(1)|m+n|+|m+p|+|q+p|(2)|n-m|--3|m+p|-|-n-q|+|q-p|
(2m+n-p)(2m-n+p)==?
(2m+n-p)乘(2m-n+p)=()
(m+n)-( )=2m-p
(m+n)-( )=2m-p
(m-2n-p)(m-2n+p)-(m+2n+p)(m+2n+p)
已知m、n、p满足|2m|+m=0,|n|=n,p|p|=1.化简:|n|-|m-p-1|+|p+n|-|2n+1|.已知m、n、p满足|2m|+m=0,|n|=n,p|p|=1.化简:|n|-|m-p-1|+|p+n|-|2n+1|.
(m+n)?【m-(n+p)】=2m-p 可以填 - 或 + 应该填什么?为什么?
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