Đáp án:

\(\begin{array}{l}
1,\\
I = \dfrac{{\sqrt x }}{{\sqrt x  + 1}}\\
2,\\
K = \dfrac{{\sqrt x  + 1}}{{x - 4}}
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
I = \dfrac{{\sqrt x {{\left( {1 - x} \right)}^2}}}{{1 + \sqrt x }}:\left[ {\left( {\dfrac{{1 - x\sqrt x }}{{1 - \sqrt x }} + \sqrt x } \right).\left( {\dfrac{{1 + x\sqrt x }}{{1 + \sqrt x }} - \sqrt x } \right)} \right]\\
 = \dfrac{{\sqrt x {{\left( {1 - x} \right)}^2}}}{{1 + \sqrt x }}:\left[ {\left( {\dfrac{{{1^3} - {{\sqrt x }^3}}}{{1 - \sqrt x }} + \sqrt x } \right).\left( {\dfrac{{{1^3} + {{\sqrt x }^3}}}{{1 + \sqrt x }} - \sqrt x } \right)} \right]\\
 = \dfrac{{\sqrt x {{\left( {1 - x} \right)}^2}}}{{1 + \sqrt x }}:\left[ {\left( {\dfrac{{\left( {1 - \sqrt x } \right)\left( {1 + 1.\sqrt x  + {{\sqrt x }^2}} \right)}}{{\left( {1 - \sqrt x } \right)}} + \sqrt x } \right).\left( {\dfrac{{\left( {1 + \sqrt x } \right)\left( {1 - 1.\sqrt x  + {{\sqrt x }^2}} \right)}}{{1 + \sqrt x }} - \sqrt x } \right)} \right]\\
 = \dfrac{{\sqrt x {{\left( {1 - x} \right)}^2}}}{{1 + \sqrt x }}:\left[ {\left( {1 + \sqrt x  + x + \sqrt x } \right)\left( {1 - \sqrt x  + x - \sqrt x } \right)} \right]\\
 = \dfrac{{\sqrt x {{\left( {1 - x} \right)}^2}}}{{1 + \sqrt x }}:\left[ {\left( {x + 2\sqrt x  + 1} \right)\left( {x - 2\sqrt x  + 1} \right)} \right]\\
 = \dfrac{{\sqrt x {{\left( {1 - x} \right)}^2}}}{{1 + \sqrt x }}:\left[ {{{\left( {\sqrt x  + 1} \right)}^2}.{{\left( {\sqrt x  - 1} \right)}^2}} \right]\\
 = \dfrac{{\sqrt x {{\left( {1 - x} \right)}^2}}}{{1 + \sqrt x }}:{\left[ {\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)} \right]^2}\\
 = \dfrac{{\sqrt x {{\left( {1 - x} \right)}^2}}}{{1 + \sqrt x }}:{\left( {x - 1} \right)^2}\\
 = \dfrac{{\sqrt x {{\left( {x - 1} \right)}^2}}}{{1 + \sqrt x }}.\dfrac{1}{{{{\left( {x - 1} \right)}^2}}}\\
 = \dfrac{{\sqrt x }}{{\sqrt x  + 1}}\\
2,\\
K = \left( {\dfrac{{\sqrt x  + 2}}{{x - 5\sqrt x  + 6}} - \dfrac{{\sqrt x  + 3}}{{2 - \sqrt x }} - \dfrac{{\sqrt x  + 2}}{{\sqrt x  - 3}}} \right):\left( {2 - \dfrac{{\sqrt x }}{{\sqrt x  + 1}}} \right)\\
 = \left( {\dfrac{{\sqrt x  + 2}}{{\left( {x - 2\sqrt x } \right) + \left( { - 3\sqrt x  + 6} \right)}} + \dfrac{{\sqrt x  + 3}}{{\sqrt x  - 2}} - \dfrac{{\sqrt x  + 2}}{{\sqrt x  - 3}}} \right):\dfrac{{2\left( {\sqrt x  + 1} \right) - \sqrt x }}{{\sqrt x  + 1}}\\
 = \left( {\dfrac{{\sqrt x  + 2}}{{\sqrt x \left( {\sqrt x  - 2} \right) - 3\left( {\sqrt x  - 2} \right)}} + \dfrac{{\sqrt x  + 3}}{{\sqrt x  - 2}} - \dfrac{{\sqrt x  + 2}}{{\sqrt x  - 3}}} \right):\dfrac{{2\sqrt x  + 2 - \sqrt x }}{{\sqrt x  + 1}}\\
 = \left( {\dfrac{{\sqrt x  + 2}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  - 3} \right)}} + \dfrac{{\sqrt x  + 3}}{{\sqrt x  - 2}} - \dfrac{{\sqrt x  + 2}}{{\sqrt x  - 3}}} \right):\dfrac{{\sqrt x  + 2}}{{\sqrt x  + 1}}\\
 = \dfrac{{\left( {\sqrt x  + 2} \right) + \left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right) - \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  + 1}}\\
 = \dfrac{{\sqrt x  + 2 + \left( {{{\sqrt x }^2} - {3^2}} \right) - \left( {{{\sqrt x }^2} - {2^2}} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  - 3} \right)}}:\dfrac{{\sqrt x  + 2}}{{\sqrt x  + 1}}\\
 = \dfrac{{\sqrt x  + 2 + x - 9 - x + 4}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  - 3} \right)}}:\dfrac{{\sqrt x  + 2}}{{\sqrt x  + 1}}\\
 = \dfrac{{\sqrt x  - 3}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  - 3} \right)}}:\dfrac{{\sqrt x  + 2}}{{\sqrt x  + 1}}\\
 = \dfrac{1}{{\sqrt x  - 2}}.\dfrac{{\sqrt x  + 1}}{{\sqrt x  + 2}}\\
 = \dfrac{{\sqrt x  + 1}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{\sqrt x  + 1}}{{x - 4}}
\end{array}\)