Đáp án:

\(\begin{array}{l}
a,\\
P = a - b\\
b,\\
P =  - \sqrt 2 
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
P = \left( {\dfrac{{a\sqrt a  + b\sqrt b }}{{\sqrt a  + \sqrt b }} - \dfrac{{a\sqrt b  - b\sqrt a }}{{\sqrt a  - \sqrt b }}} \right):\dfrac{{\sqrt a  - \sqrt b }}{{\sqrt a  + \sqrt b }}\\
 = \left( {\dfrac{{{{\sqrt a }^3} + {{\sqrt b }^3}}}{{\sqrt a  + \sqrt b }} - \dfrac{{{{\sqrt a }^2}.\sqrt b  - {{\sqrt b }^2}.\sqrt a }}{{\sqrt a  - \sqrt b }}} \right):\dfrac{{\sqrt a  - \sqrt b }}{{\sqrt a  + \sqrt b }}\\
 = \left( {\dfrac{{\left( {\sqrt a  + \sqrt b } \right)\left( {{{\sqrt a }^2} - \sqrt a .\sqrt b  + {{\sqrt b }^2}} \right)}}{{\sqrt a  + \sqrt b }} - \dfrac{{\sqrt a .\sqrt b .\left( {\sqrt a  - \sqrt b } \right)}}{{\sqrt a  - \sqrt b }}} \right):\dfrac{{\sqrt a  - \sqrt b }}{{\sqrt a  + \sqrt b }}\\
 = \left( {\dfrac{{\left( {\sqrt a  + \sqrt b } \right)\left( {a - \sqrt {ab}  + b} \right)}}{{\sqrt a  + \sqrt b }} - \dfrac{{\sqrt {ab} \left( {\sqrt a  - \sqrt b } \right)}}{{\sqrt a  - \sqrt b }}} \right):\dfrac{{\sqrt a  - \sqrt b }}{{\sqrt a  + \sqrt b }}\\
 = \left[ {\left( {a - \sqrt {ab}  + b} \right) - \sqrt {ab} } \right]:\dfrac{{\sqrt a  - \sqrt b }}{{\sqrt a  + \sqrt b }}\\
 = \left( {a - 2\sqrt {ab}  + b} \right):\dfrac{{\sqrt a  - \sqrt b }}{{\sqrt a  + \sqrt b }}\\
 = \left( {{{\sqrt a }^2} - 2\sqrt a .\sqrt b  + {{\sqrt b }^2}} \right):\dfrac{{\sqrt a  - \sqrt b }}{{\sqrt a  + \sqrt b }}\\
 = {\left( {\sqrt a  - \sqrt b } \right)^2}.\dfrac{{\sqrt a  + \sqrt b }}{{\sqrt a  - \sqrt b }}\\
 = \left( {\sqrt a  - \sqrt b } \right)\left( {\sqrt a  + \sqrt b } \right)\\
 = {\sqrt a ^2} - {\sqrt b ^2}\\
 = a - b\\
b,\\
a = \sqrt {2 - \sqrt 3 }  = \sqrt {\dfrac{1}{2}.\left( {4 - 2\sqrt 3 } \right)}  = \sqrt {\dfrac{1}{2}} .\sqrt {4 - 2\sqrt 3 } \\
 = \dfrac{1}{{\sqrt 2 }}.\sqrt {3 - 2\sqrt 3  + 1}  = \dfrac{1}{{\sqrt 2 }}.\sqrt {{{\sqrt 3 }^2} - 2.\sqrt 3 .1 + {1^2}} \\
 = \dfrac{1}{{\sqrt 2 }}.\sqrt {{{\left( {\sqrt 3  - 1} \right)}^2}}  = \dfrac{1}{{\sqrt 2 }}.\left| {\sqrt 3  - 1} \right| = \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt 3  - 1} \right)\\
b = \sqrt {2 + \sqrt 3 }  = \sqrt {\dfrac{1}{2}.\left( {4 + 2\sqrt 3 } \right)}  = \sqrt {\dfrac{1}{2}} .\sqrt {4 + 2\sqrt 3 } \\
 = \dfrac{1}{{\sqrt 2 }}.\sqrt {3 + 2\sqrt 3  + 1}  = \dfrac{1}{{\sqrt 2 }}.\sqrt {{{\sqrt 3 }^2} + 2.\sqrt 3 .1 + {1^2}} \\
 = \dfrac{1}{{\sqrt 2 }}.\sqrt {{{\left( {\sqrt 3  + 1} \right)}^2}}  = \dfrac{1}{{\sqrt 2 }}.\left| {\sqrt 3  + 1} \right| = \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt 3  + 1} \right)\\
 \Rightarrow P = a - b = \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt 3  - 1} \right) - \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt 3  + 1} \right)\\
 = \dfrac{1}{{\sqrt 2 }}.\left[ {\left( {\sqrt 3  - 1} \right) - \left( {\sqrt 3  + 1} \right)} \right]\\
 = \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt 3  - 1 - \sqrt 3  - 1} \right)\\
 = \dfrac{1}{{\sqrt 2 }}.\left( { - 2} \right)\\
 =  - \sqrt 2 
\end{array}\)