Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\dfrac{{\sqrt x - 2}}{{\sqrt x - 1}} - \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} + \dfrac{{6\sqrt x - 8}}{{x - 4\sqrt x + 3}}\\
= \dfrac{{\sqrt x - 2}}{{\sqrt x - 1}} - \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} + \dfrac{{6\sqrt x - 8}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right) - \left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) + 6\sqrt x - 8}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\left( {x - 5\sqrt x + 6} \right) - \left( {x - 1} \right) + 6\sqrt x - 8}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - 5\sqrt x + 6 - x + 1 + 6\sqrt x - 8}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{1}{{\sqrt x - 3}}\\
b,\\
\dfrac{{2\sqrt x - 9}}{{x - 5\sqrt x + 6}} - \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}} - \dfrac{{2.\left( {\sqrt x + 1} \right)}}{{3 - \sqrt x }}\\
= \dfrac{{2\sqrt x - 9}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}} - \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}} + \dfrac{{2.\left( {\sqrt x + 1} \right)}}{{\sqrt x - 3}}\\
= \dfrac{{\left( {2\sqrt x - 9} \right) - \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) + 2.\left( {\sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{2\sqrt x - 9 - \left( {x - 9} \right) + 2.\left( {x - \sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{2\sqrt x - 9 - x + 9 + 2x - 2\sqrt x - 4}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - 4}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x + 2}}{{\sqrt x - 3}}
\end{array}\)
