n)p(m

（m-2n-p)(m-2n+p)-(m+2n+p)(m+2n+p)（m-2n-p)(m-2n+p)-(m+2n+p)(m+2n+p)=[（m-2n）-p][(m-2n)+p]-[(m+2n)+p][(m+2n)+p]=(m-2n)^2-p^

化简,1.[(m+n-p)(m+p+n)-(m+n)²]/(-p),2.[(m+n)(m-n)-(m-n)²+2n(m-n)]/4n

a(m-n)^p*(n-m)^q*(m-n)^q*(n-m)^p等于什么?a(m-n)^p*(n-m)^q*(m-n)^q*(n-m)^p=a(m-n)^p*[-(m-n)]^q*(m-n)^q*[-(m-n)]^p=a(m-n)^p*(-

化简(-m+2n-p)(-p-2n+m)4n^2-m^2+p^2+4mn

(m-n+p)(m+n-p)等于多少(m-n+p)(m+n-p)=[m-(n-p)][m+(n-p)]=m²-(n-p)²=m²-n²+2np-p²(m-n+p)(m+n-p)=m²

(m-2n+3p)(m+3n+3p)原式=m*m+m*3n+m*3p-2n*m-2n*3n-2n*3p+3p*m+3p*3n+3p*3p=m^2+3mn+3mp-2mn-6n^2-6np+3mp+6pn+9p^2=m^2-6n^2+9p^2

(m+n)(p+q)-(m+n)(p-q)(m+n)(p+q)-(m+n)(p-q)=(m+n)(p+q-p+q)=2q(m+n)2q(m+n)提取（m+n),(m+n)(p+q)-(m+n)(p-q)=(m+n)(p+q-p+q)=(m+

(3m-n-p)(3m+n-P)(3m-n-p)(3m+n-P)=[(3m-p)-n][(3m-P)+n]=(3m-p)^2-n^2=9m^2-6mp+p^2-n^2=[3m-p]^2-n^2=9m^2-6mp+p^2-n^2

(2m+n-p)(2m-n+p)=[2m+(n-p)][2m-(n-p)]=4m²-(n-p)²=4m²-n²+2pn-p²(2m+(n-p))(2m-(n-p))=2m*2m-(n-p)(

(2m+n-p)(2m-n-p)等于什么(2m+n-p)(2m-n-p)=[(2m-p)+n][(2m-p)-n]=(2m-p)²-n²=4m²-4mp+p²-n²对我的回答有不明白的可以追

（2m+n-p）（2m+p-n）原式=（2m)²+(n-p)²

(3m+n-p)(3m-n+p)(3m+n-p)(3m-n+p)=[3m+(n-p)][3m-(n-p)]=(3m)²-(n-p)²=9m²-(n²-2np+p²)=9m²-n&s

计算(-3m+n-p)(3m-n-p)(3m-n-p)(3m+n-P)=[(3m-p)-n][(3m-P)+n]=(3m-p)^2-n^2=9m^2-6mp+p^2-n^2原式=-（3m-n+p）（3m-n-p）=-【（3m-n）+p】【（

(m+n+p-q)(m-n-p-q)怎么算(m+n+p-q)(m-n-p-q)=[(m-q)+(n+p)][(m-q)-(n+p)]=(m-q)^2-(n+p)^2=.下面你应该算了吧,O(∩_∩)O哈哈~这道题目需要我们掌握的是每两项在一

(m-2n-p)(-m-2n-p)(m-2n-p)(-m-2n-p)=(2n+p)²-m²=4n²+4np+p²-m²此题如仍有疑问,欢迎追问!

计算(p+q-m-n)(p-q-m+n)(p+q-m-n)(p-q-m+n)=[(p-m)+(q-n)][(p-m)-(q-n)]=(p-m)^2-(q-n)^2=[(p-m)+(q-n)][(p-m)-(q-n)]然后利用平方差公式计算=

化简:(2m+n-p)*(2m-n+p)(2m+n-p)*(2m-n+p)=(2m)^2-(n-p)^2=4m^2-n^2-p^2+2np

(m+n)(p+q)-(m+n)(p-q)=(m+n)(p+q)-(m+n)(p-q)=（m+n)(p+q-P+q)=(m+n)×2q=2q(m+n)提取公因式(mn)，(mn)(pq-pq)=(mn)×2q=2qm2qn

(-2m-3n-p)(2m-3n-p)原式=-[2m+(3n+p)]·[2m-(3n+p)]=-4m²+(3n+p)²=9n²+p²+6np-4m²化简？

因式分解（m+n)(p+q）-（n-m)(p-q)(m+n)(p+q)-(n-m)(p-q)=mp+mq+np+nq-(np-nq-mp+mq)=mp+mq+np+nq-np+nq+mp-mq=mp+nq+nq+mp=2mp+2nq=2(m