若(1-m)² mn-2=-c²-搜答案-大师作文网

(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,

（9-2mn+2m+3n)-(3mn+2n-2m)-(m+4n+mn)=9-2mn+2m+3n-3mn-2n+2m-m-4n-mn=9-6mn+3m-3n=9-6mn+3(m-n)=9-6×（-1）+

(-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)=-

原式=-2mn+2m+3n-3mn-2n+2m-m-4n-mn=3（m-n）-6mn=3×2-6×1=0．

=[m(m+n)／(m^2-n^2)-n(m-n)／(m^2-n^2)+2mn／(m^2-n^2)]÷(m^2+mn)／m=(m^2+mn-mn+n^2+2mn)／(m+n)(m-n)÷(m+n)=(

先合并同类项,得3(m-n)-6mn+9,代入已知数据,有结果27

因为三角形AED相似于三角形MNC所以AD/NC=AE/MC=DE/MN或AD/MC=AE/NC=DE/MN(1)若AD/NC=AE/MC=DE/MN则因为正方形ABCD的边长为2,AE=EB.所以D

(-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)=-

（9-2mn+2m+3n)-(m+4n+mn）=9-2mn+2m+3m-m-4n-mn=9-mn+(m-n)=9-(-1)+4=14

m=3,n＝-1/3所以选D

作A点关于直线MN的对称点A′，连接A′B交MN于C，则AC+BC=A′C+BC=A′B，A′B就是AC+BC的最小值；延长BN使ND=A′M，连接A′D，∵AM⊥MN，BN⊥MN，∴AA′∥BD，∴

M-N=3X的2次方-5X+2-2X的2次方+5X-1=x²+1>0M>N答案：A.M>N再问：.效率吗再答：绝对正确

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

(-n/m)÷n/(m^2-m)=(-n/m)*(m^2-m)/n=-(m-1)=1-m选B

∵m+n=5，mn=-6，∴(m+n)2 −4mn=52-4×（-6）=25+24=49，故答案为：49．

-2(mn-3m平方)-{m平方-5(mn-m平方)+2mn},其中m=1,n=-2=-2mn+6m^2-m^2+5(mn-m^2)-2mn=-2mn+6m^2-m^2+5mn-5m^2-2mn=mn

已知mn=-1,m-n=4则(-2mn+m+n)-(3mn+5n-5m)-(m+4n-3mn)=-2mn+m+n-3mn-5n+5m-m-4n+3mn=-2mn+5m-8n=2+20-3n=22-3n

(N-M)²-4MN=5²-4*(-2)=25+8=33

分式m-2n/m^2-2mn=1/()

(m-2n)/(m^2-2mn)=(m-2n)/[m(m-2n)]=1/m

|2m+n|+（n-2)²=0∴2m+n=0n-2=0∴m=-1n=2∴mn=-2选D