4n(m-2)
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(m-n)2(n-m)7
您好:(m+4n)(m-n)-6(m+2n)(m-3n)=m²+3mn-4n²-6m²+6mn+36n²=-5m²+9mn+32n²不明白,
一项一项通分,分母用平方差公式合并,分子进行加减,很简单就能算出来.别图省事,只要动手就能算出来的.
(3m-4n)(4n+3m)-(2m-n)(2m+n)=(9m²-16n²)-(4m²-n²)=9m²-16n²-4m²+n
2(m+n)²-(m+n)+4(m+n)-(m+n)²+3(m+n)²=4(m+n)²+3(m+n)
4(m+n)^2·(-m-n)^3-(m+n)(-m-n)^4+5(m+n)^5=-4(m+n)^2·(m+n)^3-(m+n)(m+n)^4+5(m+n)^5=-4(m+n)^5-(m+n)^5+5
4(m+n)^2乘(-m-n)^3-(m+n)(-m-n)^4+5(m+n)^5=-4(m+n)^5-(m+n)^5+5(m+n)^5=0
4n^2-(m+n)^2=(2n+m+n)(2n-m-n)=(3n+m)(n-m)
[(m+n)(m-n)-(m-n)^2+2n(m-n)]÷4n=[m²-n²-m²+2mn-n²+2nm-2n²]÷4n=4mn÷4n=m
=(m²+3mn-4n²)-6(m²-mn-6n²)=m²+3mn-4n²-6m²+6mn+36n²=-5m²
原式=(m²-n²-m²+2mn-n²+2mn-2n²)÷4n=(-4n²+4mn)÷4n=-4n²÷4n+4mn÷4n=-n+m
(2m+3n)(2m-n)-4n(2m-n)=(2m-n)(2m+3n-4n)=2(2m-n)(m-2n)
设m-n为a(a^2*a^3)^2/a^4=a^6即(m-n)^6
4m^2(m-n)+4n(n-m)=4m^2(m-n)-4n(m-n)=4(m-n)(m^2-n)
=4*(m+n)^2*(-1)*(m+n)^3-(m+n)*(-1)^4*(m+n)^4+5(m+n)^5=-4(m+n)^5-(m+n)^5+5(m+n)^5=0
由题意知,m≥0,n≥0,所以,左边≥(2√mn)²/2+(m+n)/4=2mn+(m+n)/4=(mn+m/4)+(mn+n/4)≥2√(mn•m/4)+2√(mn•
=(m^2-n^2)(-m^2-n^2)-(4m^2-n^2)(4m^2+n^2)=-(m^4-n^4)-(16m^4-n^4)=-(1^4-(-2)^4)-(16*1^4-(-2)^4)=15
这是要化简么...原式=m^2+3mn-4n^2-6(m^2-mn-6n^2)=-5m^2+9mn+32n^2
(3m-4n)(4n+3m)-(2m-n)(2m+2)=9m^2-16n^2-4m^2-4m+2mn+2n=5m^2-16n^2-4m+2mn+2n再问:计算:(4a-b^2)^2再答:(4a-b^2
原式=2(m+n)^5÷[2(m+n)^3]-3(m+n)^4÷[2(m+n)^3]+(-m-n)^3÷[2(m+n)^3]=(m+n)^2-3(m+n)/2-1/2再问:��Ҳ�㵽��һ���ˣ��