若m n=3则2m² 4mn
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m+n=2mn=-42(mn-3m)-3(2n-mn)=2mn-6m-6n+3mn=5mn-6(m+n)=-20-12=-32
(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,
原式=-2mn+2m+3n-3mn-2n+2m-m-4n-mn=3(m-n)-6mn=3×2-6×1=0.
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
3(m^2+mn)-4(mn-2m^2n)+mn=3m²+3mn-4mn+2m²n+mn=3m²+2m²n=m²(3+2n)
先合并同类项,得3(m-n)-6mn+9,代入已知数据,有结果27
再问:不理解为什么-6m-6n可以变成-6*(-2)再答: 再答:懂吗再问:不明白再答:把数字待人你计算的柿子再问:懂了
(-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
-2mn+2m+3n-3mn-2n+2m-4n-m-mn=-6mn+3m-3n=-6mn+3(m-n)=6+9=15
∵m+n=5,mn=-6,∴(m+n)2 −4mn=52-4×(-6)=25+24=49,故答案为:49.
m+n=-2mn=-4则2(mn-3m)-3(2n-mn)的值=2mn-6m-6n+3mn=5mn-6(m+n)=5*(-4)-6(-2)=-20+12=-8
原式=-2mn+2m+3n-3mn-2n+2n-m-4n-mn=-6mn+m-n=-6×2+4=-8
2[mn+(-3m)]-3(2n-mn)=2mn-6m-6n+3mn=5mn-6m-6n=5mn-6(m+n)m+n=2mn=-3=-15-12=-27
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
3(m^2n+mn)-4(mn-2m^2n)+mn.=3m^2n+3mn-4mn+8m^2n+mn.=11m^2n.
(-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
2(mn-3m)-3(2n-mn)=2mn-6m-6n+3mn=2mn+3mn-6(m+n)=32
m^2-mn=14mn-3n^2=-2m^2+3mn-3n^2=(m^2-mn)+3mn-3n^2+mn=1+4mn-3n^2=1-2=-1