\(Y=\dfrac{x^4+x^2+5}{x^4+2x^2+1}\\ =\dfrac{x^4+2x^2-x^2+1-1+5}{x^4+2x^2+1}\\ =\dfrac{\left(x^4+2x^2+1\right)-\left(x^2+1\right)+5}{x^4+2x^2+1}\\ =\dfrac{\left(x^4+2x^2+1\right)}{x^4+2x^2+1}-\dfrac{x^2+1}{\left(x^2+1\right)^2}+\dfrac{5}{\left(x^2+1\right)^2}\\ =1-\dfrac{1}{x^2+1}+\dfrac{5}{\left(x^2+1\right)^2}\)
Đặt \(\dfrac{1}{x^2+1}=t\)
\(\Rightarrow Y=1-t+5t^2\\ =5t^2-t+\dfrac{1}{20}+\dfrac{19}{20}\\ =\left(5t^2-t+\dfrac{1}{20}\right)+\dfrac{19}{20}\\ =5\left(t^2-\dfrac{1}{5}t+\dfrac{1}{100}\right)+\dfrac{19}{20}\\ =5\left(t-\dfrac{1}{10}\right)^2+\dfrac{19}{20}\)
Do \(5\left(t-\dfrac{1}{10}\right)^2\ge0\forall x;t\)
\(\Rightarrow5\left(t-\dfrac{1}{10}\right)^2+\dfrac{19}{20}\ge\dfrac{19}{20}\forall x;t\)
Dấu "=" xảy ra khi:
\(5\left(t-\dfrac{1}{10}\right)^2=0\\ \Leftrightarrow t-\dfrac{1}{10}=0\\ \Leftrightarrow t=\dfrac{1}{10}\\ \Leftrightarrow\dfrac{1}{x^2+1}=\dfrac{1}{10}\\ \Leftrightarrow x^2+1=10\\ \Leftrightarrow x^2=9\\ \Leftrightarrow x=\pm3\)
Vậy \(Y_{Min}=\dfrac{19}{20}\) khi \(x=\pm3\)