(1)两式相减,m²-n²=n-m(m+n)(m-n)=n-m=-(m-n)因m≠n,故m-n≠0,故可以约分得:m+n=-1(2)m=-(n+1)与m²=n+3联立方程消去m,求解n,再解出m,最后带入求值
m²=n+3n²=m+3两式相减得m²-n²=n-m(m-n)(m+n)=-(m-n)因为m-n不等于0,两边同时除以(m-n)m+n=-1m³-2mn+n³=m³-mn+n³-mn=m(m²-n)+n(n²-m)=3m+3n=3(m+n)=-3