Đáp án:

\(\begin{array}{l}
1,\\
a,\\
\sqrt {1 - a} .\left( {1 + \sqrt {1 + a} } \right)\\
b,\\
\left( {\sqrt a  - \sqrt b } \right){\left( {\sqrt a  + \sqrt b } \right)^2}\\
c,\\
\left( {\sqrt x  - \sqrt y } \right).\left( {\sqrt x  + \sqrt y  + y} \right)\\
2,\\
a,\\
P = \dfrac{1}{2}\\
b,\\
Q =  - 2
\end{array}\)

Giải thích các bước giải:

 Ta có:

1,

\(\begin{array}{l}
a,\\
\sqrt {1 - a}  + \sqrt {1 - {a^2}} \\
 = \sqrt {1 - a}  + \sqrt {{1^2} - {a^2}} \\
 = \sqrt {1 - a}  + \sqrt {\left( {1 - a} \right).\left( {1 + a} \right)} \\
 = \sqrt {1 - a} .\left( {1 + \sqrt {1 + a} } \right)\\
b,\\
\sqrt {{a^3}}  - \sqrt {{b^3}}  + \sqrt {{a^2}b}  - \sqrt {a{b^2}} \\
 = \sqrt {{a^2}.a}  - \sqrt {{b^2}.b}  + \sqrt {{a^2}.b}  - \sqrt {{b^2}.a} \\
 = a\sqrt a  - b\sqrt b  + a\sqrt b  - b\sqrt a \\
 = \left( {a\sqrt a  - b\sqrt a } \right) + \left( {a\sqrt b  - b\sqrt b } \right)\\
 = \sqrt a .\left( {a - b} \right) + \sqrt b .\left( {a - b} \right)\\
 = \left( {a - b} \right).\left( {\sqrt a  + \sqrt b } \right)\\
 = \left( {{{\sqrt a }^2} - {{\sqrt b }^2}} \right).\left( {\sqrt a  + \sqrt b } \right)\\
 = \left( {\sqrt a  - \sqrt b } \right).\left( {\sqrt a  + \sqrt b } \right)\left( {\sqrt a  + \sqrt b } \right)\\
 = \left( {\sqrt a  - \sqrt b } \right){\left( {\sqrt a  + \sqrt b } \right)^2}\\
c,\\
x - y + \sqrt {x{y^2}}  - \sqrt {{y^3}} \\
 = \left( {{{\sqrt x }^2} - {{\sqrt y }^2}} \right) + \sqrt {{y^2}.x}  - \sqrt {{y^2}.y} \\
 = \left( {\sqrt x  - \sqrt y } \right)\left( {\sqrt x  + \sqrt y } \right) + y\sqrt x  - y\sqrt y \\
 = \left( {\sqrt x  - \sqrt y } \right)\left( {\sqrt x  + \sqrt y } \right) + y.\left( {\sqrt x  - \sqrt y } \right)\\
 = \left( {\sqrt x  - \sqrt y } \right).\left( {\sqrt x  + \sqrt y  + y} \right)\\
2,\\
a,\\
P = \left( {\dfrac{2}{{\sqrt 3  - 1}} + \dfrac{3}{{\sqrt 3  - 2}} + \dfrac{{15}}{{3 - \sqrt 3 }}} \right).\dfrac{1}{{\sqrt 3  + 5}}\\
 = \left( {\dfrac{{2.\left( {\sqrt 3  + 1} \right)}}{{\left( {\sqrt 3  - 1} \right)\left( {\sqrt 3  + 1} \right)}} + \dfrac{{3.\left( {\sqrt 3  + 2} \right)}}{{\left( {\sqrt 3  - 2} \right)\left( {\sqrt 3  + 2} \right)}} + \dfrac{{15.\left( {3 + \sqrt 3 } \right)}}{{\left( {3 - \sqrt 3 } \right)\left( {3 + \sqrt 3 } \right)}}} \right).\dfrac{1}{{\sqrt 3  + 5}}\\
 = \left( {\dfrac{{2.\left( {\sqrt 3  + 1} \right)}}{{{{\sqrt 3 }^2} - {1^2}}} + \dfrac{{3.\left( {\sqrt 3  + 2} \right)}}{{{{\sqrt 3 }^2} - {2^2}}} + \dfrac{{15.\left( {3 + \sqrt 3 } \right)}}{{{3^2} - {{\sqrt 3 }^2}}}} \right).\dfrac{1}{{\sqrt 3  + 5}}\\
 = \left( {\dfrac{{2.\left( {\sqrt 3  + 1} \right)}}{{3 - 1}} + \dfrac{{3.\left( {\sqrt 3  + 2} \right)}}{{3 - 4}} + \dfrac{{15.\left( {3 + \sqrt 3 } \right)}}{{9 - 3}}} \right).\dfrac{1}{{\sqrt 3  + 5}}\\
 = \left( {\dfrac{{2.\left( {\sqrt 3  + 1} \right)}}{2} + \dfrac{{3.\left( {\sqrt 3  + 2} \right)}}{{ - 1}} + \dfrac{{15.\left( {3 + \sqrt 3 } \right)}}{6}} \right).\dfrac{1}{{\sqrt 3  + 5}}\\
 = \left( {\left( {\sqrt 3  + 1} \right) - 3.\left( {\sqrt 3  + 2} \right) + \dfrac{5}{2}.\left( {3 + \sqrt 3 } \right)} \right).\dfrac{1}{{\sqrt 3  + 5}}\\
 = \left( {\sqrt 3  + 1 - 3\sqrt 3  - 6 + \dfrac{{15}}{2} + \dfrac{5}{2}\sqrt 3 } \right).\dfrac{1}{{\sqrt 3  + 5}}\\
 = \left( {\dfrac{1}{2}.\sqrt 3  + \dfrac{5}{2}} \right).\dfrac{1}{{\sqrt 3  + 5}}\\
 = \dfrac{{\sqrt 3  + 5}}{2}.\dfrac{1}{{\sqrt 3  + 5}}\\
 = \dfrac{1}{2}\\
b,\\
Q = \left( {\dfrac{{\sqrt {14}  - \sqrt 7 }}{{1 - \sqrt 2 }} + \dfrac{{\sqrt {15}  - \sqrt 5 }}{{1 - \sqrt 3 }}} \right):\dfrac{1}{{\sqrt 7  - \sqrt 5 }}\\
 = \left( {\dfrac{{\sqrt 7 .\left( {\sqrt 2  - 1} \right)}}{{1 - \sqrt 2 }} + \dfrac{{\sqrt 5 .\left( {\sqrt 3  - 1} \right)}}{{1 - \sqrt 3 }}} \right):\dfrac{1}{{\sqrt 7  - \sqrt 5 }}\\
 = \left( { - \sqrt 7  - \sqrt 5 } \right):\dfrac{1}{{\sqrt 7  - \sqrt 5 }}\\
 =  - \left( {\sqrt 7  + \sqrt 5 } \right):\dfrac{1}{{\sqrt 7  - \sqrt 5 }}\\
 =  - \left( {\sqrt 7  + \sqrt 5 } \right).\left( {\sqrt 7  - \sqrt 5 } \right)\\
 =  - \left( {{{\sqrt 7 }^2} - {{\sqrt 5 }^2}} \right)\\
 =  - \left( {7 - 5} \right)\\
 =  - 2
\end{array}\)