4n^2(m-2)-6(2-m)
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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,
(2m-n)(m+n)+(m-n)²-(6m³-4m²n)÷2m=2m²+mn-n²+m²-2mn+n²-(3m²-2m
(3m-4n)(4n+3m)-(2m-n)(2m+n)=(9m²-16n²)-(4m²-n²)=9m²-16n²-4m²+n
这就是一个简单的平方差公式吧=(3m+2n)2-(2m-12n)2=(3m+2n+2m-12n)(3m+2n-2m+12n)=(5m-10n)(m+14n)
=(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
(m+2n)-6(m+2n)(2m-n)+9(n-2m)=[m+2n+3(n-2m)]^2=(5n-5m)^2=25(n-m)^2
m^2+4m+n^2-6n+13=0m^2+4m+4+n^2-6n+9=0(m+2)^2+(n-3)^2=0所以m=-2n=3m+n=-2+3=1
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)
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•
m²+²-6n+4m+13=0(m²+4m+4)+(n²-6n+9)=0(m+2)²+(n-3)²=0两非负数和为0,故(m+2)²
原式=(m-n)²+2(m-n)-6=(m+n-2)(m+n+3)
原式=(m²-mn-5mn+5n²)-6(m²-3mn+2mn-6n²)=(m²-6mn+5n²)-6(m²-mn-6n²
原式=3m(m+n)-6(m+n)(m-n)=(m+n)×【3m-6(m-n)】=(m+n)×(6n-3m)=3(m+n)(2n-m)
=(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
m-n+2n^2/(m+n)=[(m-n)(m+n)+2n^2]/(m+n)=(m^2+n^2)/(m+n)
这是要化简么...原式=m^2+3mn-4n^2-6(m^2-mn-6n^2)=-5m^2+9mn+32n^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��