已知m-2n=5,求2m-(m² 4n²) 2

应该是10吧

此题先通分,通过计算可得m^2+m^2-n^2/m^2-n^2=1+m^2/m^2-n^2=1+25/16=41/16

|2m-5|+(2m-5n+20)²=0∴2m-5=02m-5n+20=0∴m=5/2n=5(m-n)²-(2m+n)(n-m)=(m-n)²+(2m+n)(m-n)=(

(m+n)²+|m|=m绝对值项和平方项恒非负,要等式成立,m≥0,等式变为(m+n)²+m=m(m+n)²=0n=-m|2m-n-2|=02m-n-2=02m-(-m)

化简：[6m²-5m(-m+3n)+4m(-4m+5/2n)]=6m²+5m²-15mn-16m²+10mn=-5m²-5mn=-5m（m+n）；而m

m/(m+n)+m/(m-n)-n^2/(m^2-n^2)=m(m-n)/(m+n)(m-n)+m(m+n)/(m-n)(m+n)-n^2/(m^2-n^2)=（m(m-n)+m(m+n)）/(m^2

已知,m=5又4/7,n=4又3/7,可得：m+n=10,m-n=1又1/7=8/7.[-3又1/2（m+n）]^3×（m-n）×[-2（m+n）（m-n）]^2=（-7/2）^3×（m+n）^3×（

m/(m+n)+m/(m-n)-n²/((m²-n²)=(m²-mn+m²+mn-n²)/(m²-n²)=(2m&sup

m2=m+2m3=m(n+2)=mn+2mn2=m+2n3=n(m+2)=mn+2nm2-n2=(n+2)-(m+2)(m+n)(m-n)=-(m-n)m≠n则m-n≠0所以m+n=-1原式=mn+2

"m/(m+n)+m/(m-n)-n^2/(m^2-n^2)=(m/n)/(1+m/n)+(m/n)/(-1+m/n)-1/((m/n)^2-1)=(5/3)/(1+5/3)+(5/3)/(5/3-1

合并多项式,（m/m+n)+(m/m-n)-(m^2/m^2-n^2)=2m^2/(m^2-n^2)-(m^2/m^2-n^2)=m^2/(m^2-n^2)将m=5n/3代入得,（5n/3)^2/[(

[(3m+2n)(3m-2n)-(m+2n)(5m-2n)]÷(1/3)m=[9m²-4n²-5m²+2mn-10mn+4n²]÷(1/3)m=[4m²

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∵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

这种题最好能够把括号加上,不然很难分辨题目是什么合并多项式,（m/m+n)+(m/m-n)-(m^2/m^2-n^2)=2m^2/(m^2-n^2)-(m^2/m^2-n^2)=m^2/(m^2-n^

已知m=5n,则原式=(5n/(5n+n))+(5n/(5n-n))-(n^2)/(((5n)^3)-n^2)=(5/6)+(5/4)-[1/(125n-1)]=(25/12)-[1/(125n-1)

m(m^2-2mn+n)+(m-5n)=m(m-n)^2+(m-5n)带入m-n=2得4m+（m-5n)=5m-5n=5(m-n)=10

2m-n/M+2n的绝对值=32m-n/M+2n=3或-32(2m-n)/m+2n减2m-n/m+2n减3=(2m-n/m+2n)-3=0或-6

m(m²-2mn+n²）+（m-5n)=m(m-n)²+m-5n=m×2²+m-5n=4m+m-5n=5(m-n)=5×2=10

4|m+n|+4m^2+1=4m即4|m+n|+4m^2-4m+1=04|m+n|+(2m-1)^2=0根据非负数的性质：m+n=02m-1=0即m=1/2,n=-1/2(m-n)^2=[1/2-(-