Đáp án:
Xét f(x)=x^2 +2x+2016
=x^2+x+x+1+2015
=x(x+1) + (x+1) +2015
=(x+1)(x+1) +2015
= (x+1)^2 +2015
maf (x+1)^2 >= 0 vowis moij x
=> (x+1)^2 +2015 >= 2015 ( lonws hown 0 voiws moij x)
=>Dda thuwcs f(x) lowns hown 0 voiws moij x neen dda thuwcs khoong cos nghieemj( voo nghieemj)