Đáp án:
c) \(0 \le x < 36;x \ne 1\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0;x \ne 1\\
b)A = \left( {\dfrac{{x + \sqrt x - x + 2}}{{\sqrt x + 1}}} \right):\left[ {\dfrac{{\sqrt x \left( {\sqrt x - 1} \right) + \sqrt x + 4}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}} \right]\\
= \dfrac{{\sqrt x + 2}}{{\sqrt x + 1}}:\dfrac{{x - \sqrt x + \sqrt x + 4}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x + 2}}{{\sqrt x + 1}}.\dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}{{x + 4}}\\
= \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}{{x + 4}}\\
= \dfrac{{x + \sqrt x - 2}}{{x + 4}}\\
c)A < 1\\
\to \dfrac{{x + \sqrt x - 2}}{{x + 4}} < 1\\
\to \dfrac{{x + \sqrt x - 2 - x - 4}}{{x + 4}} < 0\\
\to \dfrac{{\sqrt x - 6}}{{x + 4}} < 0\\
Do:x \ge 0 \to x + 4 > 0\\
\to \sqrt x - 6 < 0\\
\to x < 36\\
\to 0 \le x < 36;x \ne 1
\end{array}\)
