Đáp án:

\(\begin{array}{l}
1,\\
K = \dfrac{{2x}}{{{x^2} - x + 1}}\\
2,\\
P = 2x - \sqrt x \\
3,\\
A = \dfrac{4}{{\sqrt a  + 3}}\\
4,\\
B = \dfrac{{x + \sqrt x  + 1}}{{\sqrt x }}\\
5,\\
C = \dfrac{{\sqrt x  - 1}}{{\sqrt x  + 1}}
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ne 1\\
x \ne  - 1
\end{array} \right.\\
K = \left( {\dfrac{1}{{x - 1}} - \dfrac{1}{{x + 1}}} \right).\dfrac{{{x^2} - 1}}{{{x^2} - x + 1}}\\
 = \dfrac{{\left( {x + 1} \right) - \left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{{x^2} - 1}}{{{x^2} - x + 1}}\\
 = \dfrac{{x + 1 - x + 1}}{{{x^2} - {1^2}}}.\dfrac{{{x^2} - 1}}{{{x^2} - x + 1}}\\
 = \dfrac{{2x}}{{{x^2} - 1}}.\dfrac{{{x^2} - 1}}{{{x^2} - x + 1}}\\
 = \dfrac{{2x}}{{{x^2} - x + 1}}\\
2,\\
DKXD:\,\,\,\,x > 0\\
P = \left( {\dfrac{{x\sqrt x }}{{\sqrt x  + 1}} + \dfrac{{{x^2}}}{{x\sqrt x  + x}}} \right).\left( {2 - \dfrac{1}{{\sqrt x }}} \right)\\
 = \left( {\dfrac{{x\sqrt x }}{{\sqrt x  + 1}} + \dfrac{{{x^2}}}{{x\left( {\sqrt x  + 1} \right)}}} \right).\dfrac{{2\sqrt x  - 1}}{{\sqrt x }}\\
 = \left( {\dfrac{{x\sqrt x }}{{\sqrt x  + 1}} + \dfrac{x}{{\sqrt x  + 1}}} \right).\dfrac{{2\sqrt x  - 1}}{{\sqrt x }}\\
 = \dfrac{{x\sqrt x  + x}}{{\sqrt x  + 1}}.\dfrac{{2\sqrt x  - 1}}{{\sqrt x }}\\
 = \dfrac{{x.\left( {\sqrt x  + 1} \right)}}{{\sqrt x  + 1}}.\dfrac{{2\sqrt x  - 1}}{{\sqrt x }}\\
 = x.\dfrac{{2\sqrt x  - 1}}{{\sqrt x }}\\
 = {\sqrt x ^2}.\dfrac{{2\sqrt x  - 1}}{{\sqrt x }}\\
 = \sqrt x .\left( {2\sqrt x  - 1} \right)\\
 = 2x - \sqrt x \\
3,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
a > 0\\
a \ne 9
\end{array} \right.\\
A = \left( {\dfrac{{\sqrt a  + 3}}{{\sqrt a  - 3}} - \dfrac{{\sqrt a  - 3}}{{\sqrt a  + 3}}} \right).\left( {\dfrac{1}{3} - \dfrac{1}{{\sqrt a }}} \right)\\
 = \dfrac{{{{\left( {\sqrt a  + 3} \right)}^2} - {{\left( {\sqrt a  - 3} \right)}^2}}}{{\left( {\sqrt a  - 3} \right)\left( {\sqrt a  + 3} \right)}}.\dfrac{{\sqrt a  - 3}}{{3\sqrt a }}\\
 = \dfrac{{\left( {{{\sqrt a }^2} + 2.\sqrt a .3 + {3^2}} \right) - \left( {{{\sqrt a }^2} - 2.\sqrt a .3 + {3^2}} \right)}}{{\left( {\sqrt a  - 3} \right)\left( {\sqrt a  + 3} \right)}}.\dfrac{{\sqrt a  - 3}}{{3\sqrt a }}\\
 = \dfrac{{\left( {a + 6\sqrt a  + 9} \right) - \left( {a - 6\sqrt a  + 9} \right)}}{{\left( {\sqrt a  - 3} \right)\left( {\sqrt a  + 3} \right)}}.\dfrac{{\sqrt a  - 3}}{{3\sqrt a }}\\
 = \dfrac{{a + 6\sqrt a  + 9 - a + 6\sqrt a  - 9}}{{\left( {\sqrt a  - 3} \right)\left( {\sqrt a  + 3} \right)}}.\dfrac{{\sqrt a  - 3}}{{3\sqrt a }}\\
 = \dfrac{{12\sqrt a }}{{\left( {\sqrt a  - 3} \right)\left( {\sqrt a  + 3} \right)}}.\dfrac{{\sqrt a  - 3}}{{3\sqrt a }}\\
 = \dfrac{4}{{\sqrt a  + 3}}\\
4,\\
DKXD:\,\,\,\,\,\,x > 0\\
B = \left( {\dfrac{1}{{\sqrt x }} + \dfrac{{\sqrt x }}{{\sqrt x  + 1}}} \right):\dfrac{{\sqrt x }}{{x + \sqrt x }}\\
 = \dfrac{{\left( {\sqrt x  + 1} \right) + {{\sqrt x }^2}}}{{\sqrt x \left( {\sqrt x  + 1} \right)}}:\dfrac{{\sqrt x }}{{{{\sqrt x }^2} + \sqrt x }}\\
 = \dfrac{{\sqrt x  + 1 + x}}{{\sqrt x \left( {\sqrt x  + 1} \right)}}:\dfrac{{\sqrt x }}{{\sqrt x \left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{x + \sqrt x  + 1}}{{\sqrt x \left( {\sqrt x  + 1} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x  + 1} \right)}}{{\sqrt x }}\\
 = \dfrac{{x + \sqrt x  + 1}}{{\sqrt x }}\\
5,\\
DKXD:\,\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right.\\
C = \dfrac{{\sqrt x }}{{\sqrt x  - 1}} + \dfrac{3}{{\sqrt x  + 1}} - \dfrac{{6\sqrt x  - 4}}{{x - 1}}\\
 = \dfrac{{\sqrt x }}{{\sqrt x  - 1}} + \dfrac{3}{{\sqrt x  + 1}} - \dfrac{{6\sqrt x  - 4}}{{{{\sqrt x }^2} - {1^2}}}\\
 = \dfrac{{\sqrt x }}{{\sqrt x  - 1}} + \dfrac{3}{{\sqrt x  + 1}} - \dfrac{{6\sqrt x  - 4}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{\sqrt x .\left( {\sqrt x  + 1} \right) + 3.\left( {\sqrt x  - 1} \right) - \left( {6\sqrt x  - 4} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{{{\sqrt x }^2} + \sqrt x  + 3\sqrt x  - 3 - 6\sqrt x  + 4}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{{{\sqrt x }^2} - 2\sqrt x  + 1}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{{{\sqrt x }^2} - 2.\sqrt x .1 + {1^2}}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{{{\left( {\sqrt x  - 1} \right)}^2}}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{\sqrt x  - 1}}{{\sqrt x  + 1}}
\end{array}\)