Đáp án:
\(\begin{array}{l}
c,\,\,\,\,\,\dfrac{{\sqrt x - 3}}{{\sqrt x + 3}}\\
d,\,\,\,\,\,\dfrac{{\sqrt x - 4}}{{\sqrt x + 1}}\\
f,\,\,\,\,\dfrac{{\sqrt x - 3}}{{\sqrt x - 2}}
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
c,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x \ne 9
\end{array} \right.\\
\dfrac{{x - 6\sqrt x + 9}}{{x - 9}} = \dfrac{{{{\sqrt x }^2} - 2.\sqrt x .3 + {3^2}}}{{{{\sqrt x }^2} - {3^2}}}\\
= \dfrac{{{{\left( {\sqrt x - 3} \right)}^2}}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}} = \dfrac{{\sqrt x - 3}}{{\sqrt x + 3}}\\
d,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right.\\
\dfrac{{x - 5\sqrt x + 4}}{{x - 1}} = \dfrac{{\left( {x - \sqrt x } \right) + \left( { - 4\sqrt x + 4} \right)}}{{{{\sqrt x }^2} - {1^2}}}\\
= \dfrac{{\left( {{{\sqrt x }^2} - \sqrt x } \right) - \left( {4\sqrt x - 4} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 1} \right) - 4.\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 4} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x - 4}}{{\sqrt x + 1}}\\
f,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x \ne 4
\end{array} \right.\\
\dfrac{{x - \sqrt x - 6}}{{x - 4}} = \dfrac{{\left( {x - 3\sqrt x } \right) + \left( {2\sqrt x - 6} \right)}}{{{{\sqrt x }^2} - {2^2}}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 3} \right) + 2\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x - 3}}{{\sqrt x - 2}}
\end{array}\)
