计算 2m^2n 3pq^2*5p^2q 4mn^2

(m-2n+3p)(m+2n+3p)=[(3p+m)+2n][(3p+m)-2n]=(3p+m)^2-(2n)^2=9p^2+6pm+m^2-4n^2

1.10.3*9.7=（10+0.3）（10-0.3）=10*10-0.3*0.3=100-0.09=99.912.(2m+n-p)(2m-n+p)=2m*2m-(n-p)(n-p)=4m²

2m/5n²p-3n/4mp²=(8m²p-15n³)/20mn²p²3y/2x+2y+2xy/x²+xy=3y/2(x+y)+2

=（2m+3n-p+2m-3n+p）（2m+3n-p-2m+3n-p）=4m（6n-2p）=24mn-8p再问：抱歉啊，那个符号可能不太明显，两个算式之间有一个减号。再答：相当于公式：A²-

原式=[（m+2n）-p]2，=（m+2n）2-2p（m+2n）+p2，=m2+4mn+4n2-2pm-4pn+p2．

原式=7m*16m^4p^2÷7m^2=(7÷7)*m*16*(m^4÷m^2)*p^2=16m*m^2*p^2=16m^3p^2

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

(m-2n+p)(m+2n-p)=[m-(2n-p)][m+(2n-p)]=m^2-(2n-p)^2=m^2-4n^2+4np-p^2（m-2n+p）（m+2n+p）=[(m+p)-2n][(m+p)

（2m+n-p)(2m-n+p)=[2m-(p-n)][2m+(p-n)]=(2m)^2-(p-n)^2=4m^2-p^2-n^2+2np不明白欢迎来追问!多谢了!再问：你确定这是最简的？再答：对啊这

∵m、n、p都是整数,∴m－n、p－m都是整数,∴｜m－n｜^3、｜p－m｜^5都是非负整数,又｜m－n｜^3＋｜p－m｜^5＝1,∴｜m－n｜、｜p－m｜只能是一者为1,另一者为0.一、当m－n＝1

已知|m-5|、|2n-3|、|4-p|应为正数,所以可得|m-5|=0、|2n-3|=0、|4-p|=0,根据上述3个式子可得m=5、n=1.5、p=4

原式=(1+2015)m/(2×2)-（2+2014）m/(2×2)=0

PrivateSubCommand1_Click()Fori=1To2000IfPrime(Sum_Y(i))=TrueAndPrime(i)=FalseThenPrintiNextEndSub'求素

m，n．p都是整数，且|m-n|3+|p-m|5=1∴|m-n|=1，p-m=0；或m-n=0，|p-m|=1①当|m-n|=1，p-m=0时p=m，|n-m|=1|p-m|+|m-n|+2|n-p|

前面括号和后面括号里是：各有（1+2009）/2=1005项去掉括号,写成1-2+3-4+5-6.+2009-2010相当于1005个-1相加,最后是-1005

(m+3m+5m+...+2009m)-(2m+4m+6m+...+2010m)=2010m/2=1005m

2010m/2=1005m(m+3m+5m+…+2011m)-(2m+4m+6m+…+2010m)=(m+3m+5m+…+2011m)-2m-4m-6m-...-2010m=m-2m+3m-4m+5m

=（m-2m)+(3m-4m)+.(一共2010/2=1005项）+2011m=-1005m+2011m=1006m

（m-2n+3p）(m+2n+3p)=m(m+2n+3p)-2n(m+2n+3p)+3p(m+2n+3p)=m²+2mn+3mp-2nm-4n²-6np+3pm+6pn+9p

∵P-Q+Q=3m2-2m-5+2m2-3m-2=5m2-5m-7,∴P+Q=5m2-5m-7+2m2-3m-2=7m2-8m-9,或直接计算P-Q+2Q得P+Q也可．