Đáp án: $\left\{ \begin{array}{l}
F = - 1\,khi:x > 2\\
F = 1\,khi:1 \le x < 2
\end{array} \right.$
Giải thích các bước giải:
$\begin{array}{l}
Dkxd:\left\{ \begin{array}{l}
x \ge 1\\
x - 2\sqrt {x - 1} > 0
\end{array} \right.\\
Xet:x - 2\sqrt {x - 1} \\
= x - 1 - 2\sqrt {x - 1} + 1\\
= {\left( {\sqrt {x - 1} } \right)^2} - 2\sqrt {x - 1} + 1\\
= {\left( {\sqrt {x - 1} - 1} \right)^2}\\
Khi:x - 2\sqrt {x - 1} > 0\\
\Leftrightarrow {\left( {\sqrt {x - 1} - 1} \right)^2} > 0\\
\Leftrightarrow \sqrt {x - 1} \# 1\\
\Leftrightarrow x - 1\# 1\\
\Leftrightarrow x\# 2\\
\Leftrightarrow Dkxd:\left\{ \begin{array}{l}
x \ge 1\\
x\# 2
\end{array} \right.\\
F = \dfrac{{1 - \sqrt {x - 1} }}{{\sqrt {x - 2\sqrt {x - 1} } }}\\
= \dfrac{{1 - \sqrt {x - 1} }}{{\sqrt {{{\left( {\sqrt {x - 1} - 1} \right)}^2}} }}\\
= \dfrac{{1 - \sqrt {x - 1} }}{{\left| {\sqrt {x - 1} - 1} \right|}}\\
+ Khi:\sqrt {x - 1} - 1 > 0\\
\Leftrightarrow \sqrt {x - 1} > 1\\
\Leftrightarrow x > 2\\
\Leftrightarrow F = \dfrac{{1 - \sqrt {x - 1} }}{{\sqrt {x - 1} - 1}} = - 1\\
+ Khi:1 \le x < 2\\
\Leftrightarrow F = \dfrac{{1 - \sqrt {x - 1} }}{{1 - \sqrt {x - 1} }} = 1\\
Vậy\,\left\{ \begin{array}{l}
F = - 1\,khi:x > 2\\
F = 1\,khi:1 \le x < 2
\end{array} \right.
\end{array}$
