(1)(m+n)^-4m(m+n)+4m^
(1)(m+n)^-4m(m+n)+4m^
证明:4/1(m*m+n*n-m-n)必为整数..m,n都是正整数...
先化简,在求值(m+n)(m-n)(-m^2-n^2)-(-2m+n)(-2m-n)(4m^2+n^2) 其中m=1,n
(m+n)²-4m(m+n)+4m²
因式分解4m^2(m-n)+4n(n-m)
[(m+n)(m-n)-(m-n)²+2n(m-n)]÷4n
(m-n)2(n-m)3(n-m)4化简
[(m-n)^2*(m-n)^3]^2/(m-n)^4
m²n(m-n)-4mn(n-m)
计算1/(m-n)-1/(m+n)-2n/(m^2+n^2)-4n^3/(m^4+n^4)-8n^7/(m^8+n^8)
已知4|m+n|+4m^2+1=4m,求根号(m-n)^2.
(3m-4n)(4n+3m)-(2m-n)(2m+n)