\(P=x^2+x+1=x^2+\frac{1}{2}2x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Ta luôn có \(\left(x+\frac{1}{2}\right)^2\ge0\) với mọi x
=> \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\) với mọi x
Dâu '' = '' xảy ra khi x = \(\frac{1}{2}\)
Pmin = \(\frac{3}{4}\) khi x= \(\frac{1}{2}\)
P=x2+x+1
=( x2+2.\(\frac{1}{2}x+\frac{1}{4}\))+1\(-\frac{1}{4}\)
=(x+\(\frac{1}{2}\))2+\(\frac{3}{4}\ge\frac{3}{4}\)
Dấu = xảy ra khi : (x+\(\frac{1}{2}\))2=0
\(\Leftrightarrow x=-\frac{1}{2}\)
vậy Min P=\(\frac{3}{4}\Leftrightarrow x=-\frac{1}{2}\)
