\(A=\dfrac{2x^2}{x^2}+\dfrac{4x}{x^2}+\dfrac{4}{x^2}=\dfrac{4}{x^2}+\dfrac{4}{x}+2=\left(\dfrac{4}{x^2}+\dfrac{4}{x}+1\right)+1=\left(\dfrac{2}{x}+1\right)^2+1\ge1\)
\(A_{min}=1\) khi \(\dfrac{2}{x}+1=0\Rightarrow x=-2\)
\(B=1-\dfrac{2}{x}+\dfrac{2012}{x^2}=2012\left(\dfrac{1}{x}-\dfrac{1}{2012}\right)^2+\dfrac{2011}{2012}\ge\dfrac{2011}{2012}\)
\(B_{min}=\dfrac{2011}{2012}\) khi \(x=2012\)
C: đặt \(x+2016=t\Rightarrow x=t-2016\)
\(C=\dfrac{t-2016}{t^2}=-\dfrac{2016}{t^2}+\dfrac{1}{t}=-2016\left(\dfrac{1}{t}-\dfrac{1}{4032}\right)^2+\dfrac{1}{8064}\le\dfrac{1}{8064}\)
\(C_{max}=\dfrac{1}{8064}\) khi \(t=4032\Rightarrow x=2016\)
\(D=\dfrac{x^2-x+1}{\left(x+1\right)^2}\), đặt \(x+1=t\Rightarrow x=t-1\)
\(D=\dfrac{\left(t-1\right)^2-\left(t-1\right)+1}{t^2}=\dfrac{t^2-3t+3}{t^2}=\dfrac{3}{t^2}-\dfrac{3}{t}+1=3\left(\dfrac{1}{t}-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\)
\(D_{min}=\dfrac{1}{4}\) khi \(t=2\Rightarrow x=1\)
