\(\frac{x^2+x+1}{x^2+2x+1}=\frac{\left(x^2+x+\frac{1}{4}\right)+\frac{3}{4}}{\left(x+1\right)^2}\)
Vậy GTNN của biểu thức là \(\frac{3}{4}\)
Đặt \(y=\frac{x^2+x+1}{x^2+2x+1}\)
\(x^2y+2xy+y=x^2+x+1\)
\(\left(1-y\right)x^2+\left(1-2y\right)x+\left(1-y\right)=0\)
\(\Delta=\left(1-2y\right)^2-4\left(1-y\right)^2\)
\(\Delta=4y-3\)
\(4y-3\ge0\)
Để đạt GTNN thì \(y=\frac{3}{4}\)
\(Min_Q=\frac{3}{4}\) tại x=1
\(Q=\frac{x^2+x+1}{x^2+2x+1}=\frac{x^2+x+1}{x^2+2x+1}-0.75+0.75=\frac{x^2+x+1-0.75x^2-1.5x-0.75}{x^2+2x+1}+0.75\) \(=\frac{0.25x^2-0.5x+0.25}{\left(x+1\right)^2}+0.75=\frac{0.25\left(x-1\right)^2}{\left(x+1\right)^2}+0.75\ge0.75\) \(minQ=0.75\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)
