Đáp án:
\(\forall x \ge 0;x \ne 1\)
Giải thích các bước giải:
\(\begin{array}{l}
B = \dfrac{{1 - x\sqrt x }}{{1 - \sqrt x }} \ge 0\\
\to \dfrac{{\left( {1 - \sqrt x } \right)\left( {x + \sqrt x + 1} \right)}}{{1 - \sqrt x }} \ge 0\\
\to x + \sqrt x + 1 \ge 0\\
\to x + 2.\sqrt x .\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4} \ge 0\\
\to {\left( {\sqrt x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge 0\\
Do:{\left( {\sqrt x + \dfrac{1}{2}} \right)^2} \ge 0\\
\to {\left( {\sqrt x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge 0\left( {ld} \right)\forall x \ge 0;x \ne 1
\end{array}\)
