Đáp án:
\(\begin{array}{l}
C = \dfrac{{ - 5\sqrt x + 2}}{{\sqrt x + 3}}\\
D = \dfrac{{5\left( {\sqrt x - 3} \right)}}{{x - 10\sqrt x - 59}}
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
C = \dfrac{{15\sqrt x - 11}}{{x + 2\sqrt x - 3}} + \dfrac{{3\sqrt x - 2}}{{1 - \sqrt x }} - \dfrac{{2\sqrt x + 3}}{{\sqrt x + 3}}\\
= \dfrac{{15\sqrt x - 11}}{{\left( {x - \sqrt x } \right) + \left( {3\sqrt x - 3} \right)}} - \dfrac{{3\sqrt x - 2}}{{\sqrt x - 1}} - \dfrac{{2\sqrt x + 3}}{{\sqrt x + 3}}\\
= \dfrac{{15\sqrt x - 11}}{{\sqrt x \left( {\sqrt x - 1} \right) + 3.\left( {\sqrt x - 1} \right)}} - \dfrac{{3\sqrt x - 2}}{{\sqrt x - 1}} - \dfrac{{2\sqrt x + 3}}{{\sqrt x + 3}}\\
= \dfrac{{15\sqrt x - 11}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}} - \dfrac{{3\sqrt x - 2}}{{\sqrt x - 1}} - \dfrac{{2\sqrt x + 3}}{{\sqrt x + 3}}\\
= \dfrac{{\left( {15\sqrt x - 11} \right) - \left( {\sqrt x + 3} \right).\left( {3\sqrt x - 2} \right) - \left( {2\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{15\sqrt x - 11 - \left( {3x + 7\sqrt x - 6} \right) - \left( {2x + \sqrt x - 3} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{15\sqrt x - 11 - 3x - 7\sqrt x + 6 - 2x - \sqrt x + 3}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{ - 5x + 7\sqrt x - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{\left( { - 5x + 5\sqrt x } \right) + \left( {2\sqrt x - 2} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{ - 5\sqrt x \left( {\sqrt x - 1} \right) + 2.\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right).\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{\left( { - 5\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right).\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{ - 5\sqrt x + 2}}{{\sqrt x + 3}}\\
D = \left( {\dfrac{{x - 5\sqrt x }}{{x - 25}} - 1} \right):\left( {\dfrac{{25 - x}}{{x + 2\sqrt x - 15}} - \dfrac{{\sqrt x + 3}}{{\sqrt x + 5}} + \dfrac{{\sqrt x + 5}}{{\sqrt x - 3}}} \right)\\
= \left( {\dfrac{{\sqrt x \left( {\sqrt x - 5} \right)}}{{\left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}} - 1} \right):\left( {\dfrac{{25 - x}}{{\left( {x - 3\sqrt x } \right) + \left( {5\sqrt x - 15} \right)}} - \dfrac{{\sqrt x + 3}}{{\sqrt x + 5}} + \dfrac{{\sqrt x + 5}}{{\sqrt x - 3}}} \right)\\
= \left( {\dfrac{{\sqrt x }}{{\sqrt x + 5}} - 1} \right):\left( {\dfrac{{25 - x}}{{\sqrt x \left( {\sqrt x - 3} \right) + 5.\left( {\sqrt x - 3} \right)}} - \dfrac{{\sqrt x + 3}}{{\sqrt x + 5}} + \dfrac{{\sqrt x + 5}}{{\sqrt x - 3}}} \right)\\
= \dfrac{{\sqrt x - \left( {\sqrt x + 5} \right)}}{{\sqrt x + 5}}:\left( {\dfrac{{25 - x}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 5} \right)}} - \dfrac{{\sqrt x + 3}}{{\sqrt x + 5}} + \dfrac{{\sqrt x + 5}}{{\sqrt x - 3}}} \right)\\
= \dfrac{{ - 5}}{{\sqrt x + 5}}:\dfrac{{\left( {25 - x} \right) - \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) + {{\left( {\sqrt x + 5} \right)}^2}}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{ - 5}}{{\sqrt x + 5}}:\dfrac{{25 - x - \left( {x - 9} \right) + \left( {x + 10\sqrt x + 25} \right)}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{ - 5}}{{\sqrt x + 5}}:\dfrac{{25 - x - x + 9 + x + 10\sqrt x + 25}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{ - 5}}{{\sqrt x + 5}}:\dfrac{{ - x + 10\sqrt x + 59}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{ - 5}}{{\sqrt x + 5}}.\dfrac{{\left( {\sqrt x + 5} \right).\left( {\sqrt x - 3} \right)}}{{ - x + 10\sqrt x + 59}}\\
= \dfrac{{ - 5.\left( {\sqrt x - 3} \right)}}{{ - x + 10\sqrt x + 59}}\\
= \dfrac{{5\left( {\sqrt x - 3} \right)}}{{x - 10\sqrt x - 59}}
\end{array}\)