计算:(m 2n)(m-2n)(m^ 4n^)
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(2m-n)/(n-m)+m/(m-n)+n/(n-m)=(2m-n)/(n-m)-m/(n-m)+n/(n-m)=(2m-n-m+n)/(n-m)=m/(n-m)欢迎采纳!
m-m3-mn2+2m2n,=m-m(m2-2mn+n2),=m-m(m-n)2,=m[1-(m-n)2],=m(1+m-n)(1-m+n).
m+2m/n-m+n/m-n-2m/n-m=(m+2m-2m)/(n-m)+n/(m-n)=m/(n-m)+n(m-n)=-m/(m-n)+n/(m-n)=-(m-n)/(m-n)=-1
(m2n)3•(-m4n)÷(-mn)2=(m6n3)•(-m4n)÷(m2n2)=(-m10n4)÷(m2n2)=-m8n2.故答案为:-m8n2
(m+n)^3-(n+m)(m-n)^2=(m+n)[(m+n)^2-(m-n)^2]=(m+n)[(m+n+m-n)(m+n-m+n)]=(m+n)(2m+2n)=2(m+n)^2
把多项式m3-m2n-mn2+n3分解因式,先提取同类项,得m2(m-n)-n2(m-n),(m-n)(m2-n2)再根据平方差公式,得(m-n)(m-n)(m+n),因为m+n=0,所以该多项式的值
2+n(m-1)/m-m(n-2)/n
2(m-n平方)
原式=2mn(m+n)(m+n-4m),=2mn(m+n)(n-3m).
∵-2xmy与3x3yn是同类项∴m=3,n=1,∴原式=m-m2n-3m+4n+2m2n-3n=m2n-2m+n,当m=3,n=1时,原式=9×1-2×3+1=4.
M+1+N+1=16M-N≠0∴M=3,N=11或M=11,N=3∴3-M2N=-96或-360
2m-n/n-m+m/m-n+n/n-m=(2m-n-m+n)/(n-m)=m/(n-m)
因为m+n=5,mn=-14,所以m2n+mn2=mn(m+n)=-14×5=-70.
mn+mn=mn(m+n)=3*5=15
原式=【(m-n)(n-m)】^2=[m(n-m)-n(n-m)]^2=[mn-m^2-n^2+mn]^2=(2mn-m^2-n^2)^2=(m-n)^4
今晚先回答的问题:雅安平安!n/(m-n)x(m^3+mn^2-2m^2n)/(n^3)÷(n^2-m^2)/(mn+n^2)=n/(m-n)xm(m-n)²/(n^3)÷(n-m)(m+n
原式=2mn(4m2n2-3mn2-3m2n2-mn2)=2mn(m2n2-4mn2)=2m3n3-8m2n3.
原式=-(m^2+2mn+n^2)=-m^2-2mn-n^2
m,n是方程x2-2010x-1=0的两个实数根∴m+n=2010,mn=-1m²n+mn²-mn=mn(m+n-1)=-1×(2010-1)=-2009
6mx2+4nxy+2x+2xy-x2+y+4=(6m-1)x2+(4n+2)xy+2x+4,由结果中不含二次项,得到6m-1=0,4n+2=0,即m=16,n=-12,则原式=6m-2n+2=1+1