Question

试说明,不论a,b为任何有理数,代数式a^2+b^2-2a+6b+11的值永远大于0

Answer 1

a^2+b^2-2a+6b+11

＝(a^2-2a+1)+(b^2+6b+9)+1

=(a-1)^2+(b+3)^2+1

不论a,b取何值,都有(a-1)^2≥0,(b+3)^2≥0

所以(a-1)^2+(b+3)^2+1＞0