mn8,mn12,求mn的值-搜答案-智慧作业帮

由于:mn/(m+n)=2则有：mn=2(m+n)则:原式=(3m+3n-5mn)/(-m-n+3mn)=[3(m+n)-5mn]/[-(m+n)+3mn]=[3(m+n)-10(m

(2mn+2m+3n)-(3mn+2n-2m)-(m+4n+mn)先去括号=2mn+2m+3n-3mn-2n+2m-m-4n-mn合并同类项=-2mn+3m-3n=-2mn+3(m-n)把m-n=2,

2(mn+m)-[-(3n-mn)-m]+mn　　=2mn+2m+3n-mn+m+mn　　=2mn+3m+2n　　=2mn+3(m+n)　　∵m+n=3,mn=-2　　∴2(mn+m)-[-(3n-m

将两式相加可得M2-N2=2,两式相减M2-2MN+N2=14

求MN的值

解题思路：根据题意，由根的判别式和不等式可求解题过程：varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/inc

解（-2mn+2m+3n)-(3mn+2n-2m)-(m+4n+mn）=-2mn+2m+3n-3mn-2n+2m-m-4n-mn=-2mn-3mn-mn+2m+2m-m+3n-2n-4n=-6mn+3

(-2mn+2m+3n)-(3mn+2n-2m)-(m+4n+mn)=-2mn+2m+3n-3mn-2n+2m-m-4n-mn=(-2mn-3mn-mn)+(2m+2m-m)+(3n-2n-4n)=-

因为m-mn=21,mn-n=15,所以：m-n=(m-mn)+(mn-n)=21+15=36m-2mn+n=(m-mn)-(mn-n)=21-15=6希望能都帮到你,追问：对不起啊.我把题发错了,m

已知m2-mn=21.mn-n2=-15两式相减得：m2-2mn+n2=36再问：m后面是2次方。n后面也是2次方再答：是的已知m^2-mn=21.mn-n^2=-15两式相减得：m^2-2mn+n^

-MN(M^2N^5-MN^3-N)=-（-6)^3+(-6)^2-(-6)=258

-2mn+2m+3n-3mn-2n+2m-4n-m-mn=-6mn+3m-3n=-6mn+3(m-n)=6+9=15

根据题意绝对值和完全平方非负所以mn-1=0m-n-2=0mn=1m-n=2（-2mn+2m+3n)-(3mn+2n-2m)-(m+4n+mn)=-2mn+2m+3n-3mn-2n+2m-m-4n-m

原式=-2mn+2m+3n-3mn-2n+2n-m-4n-mn=-6mn+m-n=-6×2+4=-8

原式=4m+3mn-4mn+6n=2(2m+3n)-mn=2×6-4=8

=(m^2-mn)+2(m^2-n^2)＝(m^2-mn)+2(m^2-mn)+2(mn-n^2)题目条件打错了,自己代入一下

∵原式=-3(2n-mn)+2(mn-3m)=-6(m+n)+5mn∵m+n=-3,mn=2∴原式=-6·-3+5·2=28

∵m2-mn=7，mn-n2=-2，∴m2-n2=（m2-mn）+（mn-n2）=7+2=9；m2-2mn+n2=（m2-mn）-（mn-n2）=7-2=5．

2（mn-3m)-3（2n-mn）=2mn-6m-6n+3mn=2mn+3mn-6(m+n)=32

∵m^2－mn＝21、mn－n^2＝15,∴两式相减,得：m^2－2mn＋n^2＝21－15＝6.

求MN的长度

解题思路：依据题意解答解题过程：varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/readq.