Đáp án:
a) \({\sqrt x - 1}\)
Giải thích các bước giải:
\(\begin{array}{l}
B5:\\
a)DK:x \ge 0;x \ne 1\\
F = \left[ {\dfrac{{2x + 1 - \sqrt x \left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}} \right].\left[ {\dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{\sqrt x + 1}} - \sqrt x } \right]\\
= \dfrac{{2x + 1 - x + \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\left( {x - \sqrt x + 1 - \sqrt x } \right)\\
= \dfrac{{x + \sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.{\left( {\sqrt x - 1} \right)^2}\\
= \sqrt x - 1\\
b)F = 3\\
\to \sqrt x - 1 = 3\\
\to x = 16\\
c)F = \dfrac{1}{3} \to \sqrt x - 1 = \dfrac{1}{3}\\
\to \sqrt x = \dfrac{4}{3}\\
\to x = \dfrac{{16}}{9}\\
d)F < x - 3\\
\to \sqrt x - 1 < x - 3\\
\to x - \sqrt x - 2 > 0\\
\to \left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right) > 0\\
\to \sqrt x - 2 > 0\left( {do:\sqrt x + 1 > 0\forall x \ge 0} \right)\\
\to x > 4\\
e)M = F.2\sqrt x = \left( {\sqrt x - 1} \right).2\sqrt x \\
= 2x - 2\sqrt x = 2x - 2.\sqrt {2x} .\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{2} - \dfrac{1}{2}\\
= {\left( {\sqrt {2x} - \dfrac{1}{{\sqrt 2 }}} \right)^2} - \dfrac{1}{2}\\
Do:{\left( {\sqrt {2x} - \dfrac{1}{{\sqrt 2 }}} \right)^2} \ge 0\forall x \ge 0\\
\to {\left( {\sqrt {2x} - \dfrac{1}{{\sqrt 2 }}} \right)^2} - \dfrac{1}{2} \ge - \dfrac{1}{2}\\
\to Min = - \dfrac{1}{2}\\
\Leftrightarrow \sqrt {2x} - \dfrac{1}{{\sqrt 2 }} = 0\\
\Leftrightarrow x = \dfrac{1}{4}
\end{array}\)
