已知m+n=2,求代数式m²(1+n)+n²(1+m)的值

问题描述:

已知m+n=2,求代数式m²(1+n)+n²(1+m)的值


m²(1+n)+n²(1+m)
=m^2+n^2+m^2n+n^2m
=m^2+n^2+mn(m+n)
=m^2+n^2+2mn
=(m+n)^2
=4

m²(1+n)+n²(1+m)=m^2+m^2n+n^2+n^2m=m^2+n^2+mn(m+n)=m^2+n^2+2mn=(m+n)^2=4

m²(1+n)+n²(1+m)=m²+m²n+n²+mn²=m²+(m²n+mn²)+n²=m²+mn(m+n)+n²=m²+2mn+n²=(m+n)²=2²=4很高兴为您解答,【the 1900】团队为您答题...