Đáp án:
b. \(x = \dfrac{1}{{121}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 0;x \ne 1\\
A = \dfrac{{15\sqrt x - 11 - \left( {3\sqrt x - 2} \right)\left( {\sqrt x + 3} \right) - \left( {2\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{15\sqrt x - 11 - 3x - 7\sqrt x + 6 - 2x - \sqrt x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{ - 5x + 7\sqrt x - 2}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\left( {2 - 5\sqrt x } \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{2 - 5\sqrt x }}{{\sqrt x + 3}}\\
b.A = \dfrac{1}{2}\\
\to \dfrac{{2 - 5\sqrt x }}{{\sqrt x + 3}} = \dfrac{1}{2}\\
\to 4 - 10\sqrt x = \sqrt x + 3\\
\to 11\sqrt x = 1\\
\to \sqrt x = \dfrac{1}{{11}}\\
\to x = \dfrac{1}{{121}}\\
c.A \le \dfrac{2}{3}\\
\to A - \dfrac{2}{3} \le 0\\
\to \dfrac{{2 - 5\sqrt x }}{{\sqrt x + 3}} - \dfrac{2}{3} \le 0\\
\to \dfrac{{6 - 15\sqrt x - 2\sqrt x - 6}}{{3\left( {\sqrt x + 3} \right)}} \le 0\\
\to - \dfrac{{17\sqrt x }}{{3\left( {\sqrt x + 3} \right)}} \le 0\\
Do:x \ge 0\\
\to \left\{ \begin{array}{l}
17\sqrt x \ge 0\\
3\left( {\sqrt x + 3} \right) > 0
\end{array} \right.\\
\to \dfrac{{17\sqrt x }}{{3\left( {\sqrt x + 3} \right)}} \ge 0\\
\to - \dfrac{{17\sqrt x }}{{3\left( {\sqrt x + 3} \right)}} \le 0\left( {ld} \right)\\
\to dpcm
\end{array}\)
