已知m n=9,mn=14,求3m的二次 3n的2次的值
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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,
把m的平方加n的平方减mn化为括号m加n括号的平方减3mn.很简单.满意请采纳
m^2-mn-n^2+3mn-3mn=(m+n)^2-3mn=81-3*14=39
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
3m-5mn+3n/(-m)+3mn-n=【3(m+n)-5mn】/【3mn-(m+n)】=【3-5mn/(m+n)】/【3mn/(m+n)-1】=【3-5x2】/【3x2-1】=-7/5
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
答:mn/(m+n)=2分子分母同除以mn得:1/(1/n+1/m)=21/m+1/n=1/2(3m-5mn+3n)/(-m+3mn-n)分子分母同除以mn得:=(3/n-5+3/m)/(-1/n+3
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+n)]/[-(m
m²-mn=15==>3m²-3mn=45(1)mn-n²=-6==>2mn-2n²=-12(2)(1)式+(2)式==>3m²-mn-2n²
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
原式=-2mn+2m+3n-3mn-2n+2n-m-4n-mn=-6mn+m-n=-6×2+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
(-m-4n-mn)-(2mn-2m-3n)-(3mn+2n-2m)=-m-4n-mn-2mn+2m+3n-3mn-2n+2m=3m-3n-6mn=3(m-n)-6mn=3×3-6×(-3)=9+18
3(2n-mn)+2(mn+3m)=6n-3mn+2mn+6m=6(m+n)-mn=6*-3-2=-20
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.