Đáp án:
$MaxN= \dfrac{1+\sqrt{5}}{2}$
Giải thích các bước giải:
Ta có:
$\dfrac{1+\sqrt{5}}{2}-N=\dfrac{1+\sqrt{5}}{2}-\dfrac{4x+1}{4x^2+1}$
$\rightarrow \dfrac{1+\sqrt{5}}{2}-N=\dfrac{(1+\sqrt{5})(4x^2+1)-2(4x+1)}{2(4x^2+1)}$
$\rightarrow \dfrac{1+\sqrt{5}}{2}-N=\dfrac{(1+\sqrt{5}).4x^2-8x+\sqrt{5}-1}{2(4x^2+1)}$
$\rightarrow \dfrac{1+\sqrt{5}}{2}-N=\dfrac{\dfrac{1+\sqrt{5}}{4}(4x+1-\sqrt{5})^2}{2(4x^2+1)}$
$\rightarrow \dfrac{1+\sqrt{5}}{2}-N\ge 0\quad \forall x$
$\rightarrow N\le \dfrac{1+\sqrt{5}}{2}$
Dấu = xảy ra khi $x=\dfrac{-1+\sqrt{5}}{4}$
