Lời giải:
ĐKXĐ: $x\neq -1$
$F=\frac{2x}{x^2+2x+1}$
$F-\frac{1}{2}=\frac{2x}{x^2+2x+1}-\frac{1}{2}=\frac{4x-x^2-2x-1}{2(x^2+2x+1)}$
$=\frac{-(x^2-2x+1)}{2(x^2+2x+1)}=\frac{-(x-1)^2}{2(x+1)^2}\leq 0$ với mọi $x\neq -1$
$\Rightarrow F\leq \frac{1}{2}$
Vậy gtln của $F$ là $\frac{1}{2}$ khi $x-1=0\Leftrightarrow x=1$
\(F=\dfrac{2x}{\left(x+1\right)^2}=\dfrac{2\left(x+1\right)-2}{\left(x+1\right)^2}=\dfrac{2}{x+1}-\dfrac{2}{\left(x+1\right)^2}\)
Đặt x + 1 = y => F = \(\dfrac{2}{y}-\dfrac{2}{y^2}\)
Đặt \(\dfrac{1}{y}=t\Rightarrow F=2t-2t^2=-2\left(t^2-t\right)=-2\left(t^2-2.t.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}\right)=-2\left(t-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\)
\(\Rightarrow F\le\dfrac{1}{2}\).Dấu "=" xảy ra khi: \(t-\dfrac{1}{2}=0\Leftrightarrow t=\dfrac{1}{2}\Leftrightarrow\dfrac{1}{y}=\dfrac{1}{2}\Leftrightarrow y=2\Leftrightarrow x+1=2\Leftrightarrow x=1\)