Đáp án:
\(\begin{array}{l}
1)B = \dfrac{1}{2}\\
2)x = \dfrac{{676}}{{25}}\\
3)0 \le x < \dfrac{{289}}{{25}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:x \ge 0\\
Thay:x = 16\\
\to B = \dfrac{{2\sqrt {16} - 5}}{{\sqrt {16} + 2}} = \dfrac{{2.4 - 5}}{{4 + 2}}\\
= \dfrac{3}{6} = \dfrac{1}{2}\\
2)B = \dfrac{3}{4}\\
\to \dfrac{{2\sqrt x - 5}}{{\sqrt x + 2}} = \dfrac{3}{4}\\
\to 8\sqrt x - 20 = 3\sqrt x + 6\\
\to 5\sqrt x = 26\\
\to \sqrt x = \dfrac{{26}}{5}\\
\to x = \dfrac{{676}}{{25}}\\
3)B < \dfrac{1}{3}\\
\to \dfrac{{2\sqrt x - 5}}{{\sqrt x + 2}} < \dfrac{1}{3}\\
\to \dfrac{{6\sqrt x - 15 - \sqrt x - 2}}{{3\left( {\sqrt x + 2} \right)}} < 0\\
\to 5\sqrt x - 17 < 0\left( {do:\sqrt x + 2 > 0\forall x \ge 0} \right)\\
\to \sqrt x < \dfrac{{17}}{5}\\
\to 0 \le x < \dfrac{{289}}{{25}}
\end{array}\)
