\(\frac{x^2+x+1}{\left(x+1\right)^2}=\frac{x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}}{\left(x+1\right)^2}=\frac{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}{\left(x+1\right)^2}\)
\(\text{Đặt }y=x+1\Rightarrow y-\frac{1}{2}=x+\frac{1}{2}\text{ ta được:}\)
\(\frac{\left(y-\frac{1}{2}\right)^2+\frac{3}{4}}{y^2}=\frac{y^2-y+\frac{1}{4}+\frac{3}{4}}{y^2}\)
\(=\frac{y^2-y+1}{y^2}=\frac{\left(y^2-y+1\right):y^2}{y^2:y^2}=1-\frac{1}{y}+\frac{1}{y^2}\)
\(=\frac{1}{4}-2.\frac{1}{y}.\frac{1}{2}+\frac{1}{y^2}+\frac{3}{4}\)
\(=\left(\frac{1}{y}-\frac{1}{2}\right)^2+\frac{3}{4}\)
Vậy GTNN của \(\frac{x^2+x+1}{\left(x+1\right)^2}\) là 3/4 tại:
\(\frac{1}{y}=\frac{1}{2}\Rightarrow y=2\Rightarrow x=y-1=2-1=1\)
\(A=\frac{x^2+x+1}{\left(x+1\right)^2}\)
Đặt \(x+1=y\Leftrightarrow x=y-1\), ta có:
\(A=\frac{\left(y-1\right)^2+y}{y^2}=\frac{y^2-2y+1+y}{y^2}=\frac{y^2-y+1}{y^2}\)
\(A=\frac{y^2}{y^2}-\frac{y}{y^2}+\frac{1}{y^2}=1-\frac{1}{y}+\frac{1}{y^2}=\left(\frac{1}{y}\right)^2-2.\frac{1}{y}.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(\frac{1}{y}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\) với mọi \(x\)
Dấu \(''=''\) xảy ra \(\Leftrightarrow\left(\frac{1}{y}-\frac{1}{2}\right)^2=0\Leftrightarrow\frac{1}{y}-\frac{1}{2}=0\Leftrightarrow\frac{1}{y}=\frac{1}{2}\Leftrightarrow y=2\)
\(\Rightarrow x=1\) (do \(x=y-1\))
Vậy, \(Min\) \(A=\frac{3}{4}\Leftrightarrow x=1\)
