原式=(2m)²+(n-p)²
(2m+n-p)( 2m+p-n)
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相关问题
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(m-2n-p)(m-2n+p)-(m+2n+p)(m+2n+p)
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M+2N-P=M+() m-2n+p=m-() m-2n-p=m-() m+2n+p=m-()
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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|.
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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|.
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计算:(2m+n-p)(2m-n+p)
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下列变形中错误的是( ) A.m 2 -(2m-n-p)=m 2 -2m+n+p B.m-n+p-q=m-(n+p-q
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化简(m+2n-3p)2-(m-2p)(m-n)
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已知m,n,p都是整数,且|m-n|+|p-m|=1,则|p-m|+|m-n|+3(n-p)2=______.
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(—2m+n—p)(2m—n—p)急求啊~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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自然数m,n,p,q 满足等式m^2+n^2=p^2+q^2 ,则m+n+p+q为