Giải thích các bước giải:
a,
ĐKXĐ: \(\left\{ \begin{array}{l}
x \ge 0\\
x + 2\sqrt x - 3 \ne 0\\
1 - \sqrt x \ne 0\\
\sqrt x + 3 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right.\)
b,
\(\begin{array}{l}
K = \frac{{15\sqrt x - 11}}{{x + 2\sqrt x - 3}} + \frac{{3\sqrt x }}{{1 - \sqrt x }} - \frac{{2\sqrt x + 3}}{{\sqrt x + 3}}\\
= \frac{{15\sqrt x - 11}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}} - \frac{{3\sqrt x \left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}} - \frac{{\left( {2\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \frac{{15\sqrt x - 11 - \left( {3x + 9\sqrt x } \right) - \left( {2x + \sqrt x - 3} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \frac{{15\sqrt x - 11 - 3x - 9\sqrt x - 2x - \sqrt x + 3}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \frac{{ - 5x + 5\sqrt x - 8}}{{x + 2\sqrt x - 3}}\\
c,\\
K = \frac{1}{2} \Leftrightarrow \frac{{ - 5x + 5\sqrt x - 8}}{{x + 2\sqrt x - 3}} = \frac{1}{2}\\
\Leftrightarrow x + 2\sqrt x - 3 = - 10x + 10\sqrt x - 16\\
\Leftrightarrow 11x - 8\sqrt x + 13 = 0\,\,\,\,\left( {ptvn} \right)
\end{array}\)
