神马作文网x^2-m^2 4mn-4n^2
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(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,
x²+2mn-m²+n²=(x+m)(x-m)+n(2m+n)
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)
4-m²n²+2mn=5-（m²n²-2mn+1）=5-（mn-1）²=（√5+mn-1）（√5-mn+1）如果本题有什么不明白可以追问,
先合并同类项,得3(m-n)-6mn+9,代入已知数据,有结果27
(-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)=-
(3m--2n)^2+24mn=9m^2--12mn+4n^2+24mn=9m^2+12mn+4n^2=(3m+2n)^2.
-2mn+2m+3n-3mn-2n+2m-4n-m-mn=-6mn+3m-3n=-6mn+3(m-n)=6+9=15
(7m²-4mn-n²)-(3m²-mn+2n²)=(7m²-4mn-n²)-3m²+mn-2n²=4m²-3
根据题意绝对值和完全平方非负所以mn-1=0m-n-2=0mn=1m-n=2（-2mn+2m+3n)-(3mn+2n-2m)-(m+4n+mn)=-2mn+2m+3n-3mn-2n+2m-m-4n-m
原式=(m²+2mn+n²-4mn)(m²-2mn+n²+4mn)+2m²n²=(m²-2mn+n²)(m²+
=4m^2n(m-n)+mn(n^2-1)=mn(4m^2-4mn+n^-1)=mn(2m-n+1)(2m-n-1)
原式=-2mn+2m+3n-3mn-2n+2n-m-4n-mn=-6mn+m-n=-6×2+4=-8
原式=3m^2-3mn+6n^2-2mn+2m^2-2n^2+5mn-4n^2+1=5m^2+1再问：（2x-3y+z)-3(x-2(-z+(x-y))-(3y-5z)再答：原式=2x-3y+z-3(
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
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
(1)(x+a)(x+b)=x²+ax+bx+ab（2）=（-12m^4n²）/4mn²=-3m^3（3）=(4x²y-2x³)/4x²=y
原式化简=mn²+2mn+4mn²-3mn=5mn²-mn注：[(mn²﹚#(2mn)]=mn²+2mn因为m#n=m+n同理后面的一样带入值运算得1