Ta có : \(B=\frac{x^2+x+1}{x^2+2x+1}=\frac{x^2+x+1}{\left(x+1\right)^2}\)
Đặt \(y=x+1\Rightarrow x=y-1\Rightarrow B=\frac{\left(y-1\right)^2+\left(y-1\right)+y}{y^2}=\frac{y^2-y+1}{y^2}=\frac{1}{y^2}-\frac{1}{y}+1\)
Đặt : \(t=\frac{1}{y}\Rightarrow B=t^2-t+1=\left(t-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Vậy \(B_{min}=\frac{3}{4}\Leftrightarrow t=\frac{1}{2}\Leftrightarrow y=2\Leftrightarrow x=1\)
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