Đáp án:
c) \(0 \le x < \dfrac{1}{9}\)
Giải thích các bước giải:
\(\begin{array}{l}
b)B = \dfrac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{x - \sqrt x + 1}}\\
P = \dfrac{A}{B} = \dfrac{{1 - \sqrt x }}{{x - \sqrt x + 1}}:\dfrac{{\sqrt x + 1}}{{x - \sqrt x + 1}}\\
= \dfrac{{1 - \sqrt x }}{{\sqrt x + 1}}\\
c)P > \dfrac{1}{2}\\
\to \dfrac{{1 - \sqrt x }}{{\sqrt x + 1}} > \dfrac{1}{2}\\
\to \dfrac{{2 - 2\sqrt x - \sqrt x - 1}}{{2\left( {\sqrt x + 1} \right)}} > 0\\
\to 1 - 3\sqrt x > 0\left( {do:\sqrt x + 1 > 0\forall x \ge 0} \right)\\
\to \dfrac{1}{9} > x\\
KL:0 \le x < \dfrac{1}{9}
\end{array}\)
