Đáp án:
\(\begin{array}{l}
1,\\
A = \dfrac{{\sqrt x - 1}}{{\sqrt x }}\\
2,\\
B = \dfrac{{\sqrt x + 2}}{{x - 16}}\\
3,\\
M = \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
A = \left( {\dfrac{1}{{x - \sqrt x }} + \dfrac{1}{{\sqrt x - 1}}} \right):\dfrac{{\sqrt x + 1}}{{{{\left( {\sqrt x - 1} \right)}^2}}}\\
= \left( {\dfrac{1}{{{{\sqrt x }^2} - \sqrt x }} + \dfrac{1}{{\sqrt x - 1}}} \right):\dfrac{{\sqrt x + 1}}{{{{\left( {\sqrt x - 1} \right)}^2}}}\\
= \left( {\dfrac{1}{{\sqrt x \left( {\sqrt x - 1} \right)}} + \dfrac{1}{{\sqrt x - 1}}} \right):\dfrac{{\sqrt x + 1}}{{{{\left( {\sqrt x - 1} \right)}^2}}}\\
= \dfrac{{1 + \sqrt x }}{{\sqrt x .\left( {\sqrt x - 1} \right)}}:\dfrac{{\sqrt x + 1}}{{{{\left( {\sqrt x - 1} \right)}^2}}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x .\left( {\sqrt x - 1} \right)}}.\dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x }}\\
2,\\
B = \left( {\dfrac{{\sqrt x }}{{\sqrt x + 4}} + \dfrac{4}{{\sqrt x - 4}}} \right):\dfrac{{x + 16}}{{\sqrt x + 2}}\\
= \dfrac{{\sqrt x .\left( {\sqrt x - 4} \right) + 4.\left( {\sqrt x + 4} \right)}}{{\left( {\sqrt x + 4} \right).\left( {\sqrt x - 4} \right)}}:\dfrac{{x + 16}}{{\sqrt x + 2}}\\
= \dfrac{{{{\sqrt x }^2} - 4\sqrt x + 4\sqrt x + 16}}{{{{\sqrt x }^2} - {4^2}}}:\dfrac{{x + 16}}{{\sqrt x + 2}}\\
= \dfrac{{x + 16}}{{x - 16}}.\dfrac{{\sqrt x + 2}}{{x + 16}}\\
= \dfrac{{\sqrt x + 2}}{{x - 16}}\\
3,\\
M = \dfrac{{2\sqrt x - 9}}{{x - 5\sqrt x + 6}} + \dfrac{{2\sqrt x + 1}}{{\sqrt x - 3}} + \dfrac{{\sqrt x + 3}}{{2 - \sqrt x }}\\
= \dfrac{{2\sqrt x - 9}}{{\left( {x - 2\sqrt x } \right) + \left( { - 3\sqrt x + 6} \right)}} + \dfrac{{2\sqrt x + 1}}{{\sqrt x - 3}} - \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}}\\
= \dfrac{{2\sqrt x - 9}}{{\sqrt x \left( {\sqrt x - 2} \right) - 3.\left( {\sqrt x - 2} \right)}} + \dfrac{{2\sqrt x + 1}}{{\sqrt x - 3}} - \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}}\\
= \dfrac{{2\sqrt x - 9}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}} + \dfrac{{2\sqrt x + 1}}{{\sqrt x - 3}} - \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}}\\
= \dfrac{{\left( {2\sqrt x - 9} \right) + \left( {2\sqrt x + 1} \right)\left( {\sqrt x - 2} \right) - \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{2\sqrt x - 9 + \left( {2x - 4\sqrt x + \sqrt x - 2} \right) - \left( {{{\sqrt x }^2} - {3^2}} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{2\sqrt x - 9 + 2x - 3\sqrt x - 2 - x + 9}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - \sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\left( {x - 2\sqrt x } \right) + \left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x .\left( {\sqrt x - 2} \right) + \left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}
\end{array}\)