Đáp án:
\(0 < A < \dfrac{8}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0;x \ne 1\\
Có:x + \sqrt x + 1 = {\left( {\sqrt x } \right)^2} + 2\sqrt x .\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4}\\
= {\left( {\sqrt x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4}\\
Do:{\left( {\sqrt x + \dfrac{1}{2}} \right)^2} > 0\forall x > 0\\
\to {\left( {\sqrt x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} > \dfrac{3}{4}\\
\to \dfrac{2}{{{{\left( {\sqrt x + \dfrac{1}{2}} \right)}^2} + \dfrac{3}{4}}} < 2:\dfrac{3}{4}\\
\to \dfrac{2}{{{{\left( {\sqrt x + \dfrac{1}{2}} \right)}^2} + \dfrac{3}{4}}} < \dfrac{8}{3}\\
Do:\left\{ \begin{array}{l}
2 > 0\\
x + \sqrt x + 1 = {\left( {\sqrt x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} > 0
\end{array} \right.\\
\to \dfrac{2}{{x + \sqrt x + 1}} > 0\\
KL:0 < \dfrac{2}{{x + \sqrt x + 1}} < \dfrac{8}{3}
\end{array}\)
