Đáp án:
\(\begin{array}{l}
13) - \dfrac{{2 - \sqrt x }}{{2\sqrt x - 3}}\\
15)\dfrac{{8\left( {\sqrt x + 1} \right)}}{{3\left( {2\sqrt x + 3} \right)}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
13)DK:x \ge 0;x \ne \left\{ {\dfrac{9}{4};4} \right\}\\
\dfrac{{\sqrt x + 2}}{{ - \sqrt x + 2}} + \dfrac{{3\sqrt x - 4}}{{2\sqrt x - 3}} + \dfrac{{ - 7\sqrt x + 10}}{{ - 2x + 7\sqrt a - 6}}\\
= \dfrac{{\left( {\sqrt x + 2} \right)\left( {2\sqrt x - 3} \right) + \left( {3\sqrt x - 4} \right)\left( {2 - \sqrt x } \right) - 7\sqrt x + 10}}{{\left( {2\sqrt x - 3} \right)\left( {2 - \sqrt x } \right)}}\\
= \dfrac{{2x + \sqrt x - 6 - 3x + 10\sqrt x - 8 - 7\sqrt x + 10}}{{\left( {2\sqrt x - 3} \right)\left( {2 - \sqrt x } \right)}}\\
= \dfrac{{ - x + 4\sqrt x - 4}}{{\left( {2\sqrt x - 3} \right)\left( {2 - \sqrt x } \right)}}\\
= \dfrac{{ - {{\left( {\sqrt x - 2} \right)}^2}}}{{\left( {2\sqrt x - 3} \right)\left( {2 - \sqrt x } \right)}}\\
= \dfrac{{ - {{\left( {2 - \sqrt x } \right)}^2}}}{{\left( {2\sqrt x - 3} \right)\left( {2 - \sqrt x } \right)}} = - \dfrac{{2 - \sqrt x }}{{2\sqrt x - 3}}\\
15)DK:x \ge 0;x \ne \dfrac{4}{9}\\
\dfrac{{ - 5\sqrt x + 4}}{{3\sqrt x - 2}} + \dfrac{{6\sqrt x + 4}}{{2\sqrt x + 3}} + \dfrac{{29\sqrt x - 28}}{{3\left( {2\sqrt x + 3} \right)\left( {3\sqrt x - 2} \right)}}\\
= \dfrac{{3\left( { - 5\sqrt x + 4} \right)\left( {2\sqrt x + 3} \right) + 3\left( {6\sqrt x + 4} \right)\left( {3\sqrt x - 2} \right) + 29\sqrt x - 28}}{{3\left( {2\sqrt x + 3} \right)\left( {3\sqrt x - 2} \right)}}\\
= \dfrac{{3\left( { - 10x - 7\sqrt x + 12} \right) + 3\left( {18x - 8} \right) + 29\sqrt x - 28}}{{3\left( {2\sqrt x + 3} \right)\left( {3\sqrt x - 2} \right)}}\\
= \dfrac{{ - 30x - 21\sqrt x + 36 + 54x - 24 + 29\sqrt x - 28}}{{3\left( {2\sqrt x + 3} \right)\left( {3\sqrt x - 2} \right)}}\\
= \dfrac{{24x + 8\sqrt x - 16}}{{3\left( {2\sqrt x + 3} \right)\left( {3\sqrt x - 2} \right)}}\\
= \dfrac{{8\left( {3\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}{{3\left( {2\sqrt x + 3} \right)\left( {3\sqrt x - 2} \right)}}\\
= \dfrac{{8\left( {\sqrt x + 1} \right)}}{{3\left( {2\sqrt x + 3} \right)}}
\end{array}\)
