Đặt \(A=\dfrac{1}{B}=\dfrac{1+x^4}{x^2}=\dfrac{1}{x^2}+x^2\ge2\sqrt{\dfrac{1}{x^2}.x^2}=2\)
Do \(A=\dfrac{1}{B}\ge2\Rightarrow B\le\dfrac{1}{2}\)
Dấu "=" xảy ra khi \(x^2=\dfrac{1}{x^2}\Leftrightarrow x=1\)
Vậy \(B_{max}=\dfrac{1}{2}\Leftrightarrow x=1\)
Easy nhở?
Cách khác nhé
\(B=\dfrac{x^2}{1+x^4}\le\dfrac{x^2}{2x^2}=\dfrac{1}{2}\)
Dấu " = " xảy ra \(\Leftrightarrow x^4+1=2x^2\Leftrightarrow\left(x^2-1\right)^2=0\Leftrightarrow x=1\)
Vậy ...