若m^2 2m n^2-6n
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mn/(3m-n)=3,则:(3m-n)/(mn)=1/3,则:2mm/(3m-n)+(6m-2n)/(mn)=2[(mn)/(3m-n)]+2×[(3m-n)/(mn)]=2×3+2×(1/3)=2
=[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)=(
3m^2-mn/9m^2-6mn+n^2=m(3m-n)/(3m-n)^2=m/(3m-n)=3n/(9n-n)=3/8
先合并同类项,得3(m-n)-6mn+9,代入已知数据,有结果27
(-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
m²-mn=15==>3m²-3mn=45(1)mn-n²=-6==>2mn-2n²=-12(2)(1)式+(2)式==>3m²-mn-2n²
算术平方根大于等于0,相加等于0,若有一个大于0,则另一个小于0,不成立.所以两个都等于0所以m-n=03m-2n=0所以m=0,n=0所以原式=2m²+5mn-4n²=0+0-0
m+n=5m-n=6则mn^2-m^2n=mn(n-m)(m+n)^2=m^2+n^2+2mn=25(m-n)^2=m^2+n^2-2mn=364mn=25-36=-11mn=-11/4mn^2-m^
m+n=-2005,mn=7由韦达定理,m,n是方程x²+2005x+7=0的根(m^2+2004m+6)(n^2+2006n+8)=(m²+2005m+7-m-1)(n²
我觉得你们老师要是出这样的题给你做,你该提醒他吃药了!
-MN(M^2N^5-MN^3-N)=-(-6)^3+(-6)^2-(-6)=258
原式=(m²+2mn+n²-4mn)(m²-2mn+n²+4mn)+2m²n²=(m²-2mn+n²)(m²+
∵mn-n=15①,m-mn=6②,∴①+②得:mn-n+m-mn=21,即m-n=21;①-②得:mn-n-m+mn=2mn-m-n=9,即-2mn+m+n=-9.故答案为21;-9
∵m+n=5,mn=-6,∴(m+n)2 −4mn=52-4×(-6)=25+24=49,故答案为:49.
=(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
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
题错了应该是m^3-m^2n-mn^2+n^3=m^3+n^3-mn(m+n)=m^3+n^3-0=(m+n)(m^2-mn+n^2)=0
(N-M)²-4MN=5²-4*(-2)=25+8=33