Đáp án:
\(\begin{array}{l}
k)1\\
l)2\\
m) - 2{x^2}\left( {y - 2} \right)
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
k)\sqrt {\sqrt 2 - 1} .\sqrt {\left( {2 - \sqrt {3 - \sqrt 2 } } \right)\left( {2 + \sqrt {3 - \sqrt 2 } } \right)} \\
= \sqrt {\sqrt 2 - 1} .\sqrt {4 - \left( {3 - \sqrt 2 } \right)} \\
= \sqrt {\sqrt 2 - 1} .\sqrt {1 + \sqrt 2 } \\
= \sqrt {2 - 1} = 1\\
l)\left( {2 - \sqrt 3 } \right).\left( {\sqrt 3 + 1} \right)\sqrt 2 .\sqrt {2 + \sqrt 3 } \\
= \left( {2 - \sqrt 3 } \right).\left( {\sqrt 3 + 1} \right)\sqrt {4 + 2\sqrt 3 } \\
= \left( {2 - \sqrt 3 } \right).\left( {\sqrt 3 + 1} \right)\sqrt {3 + 2\sqrt 3 .1 + 1} \\
= \left( {2 - \sqrt 3 } \right).\left( {\sqrt 3 + 1} \right)\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} \\
= \left( {2 - \sqrt 3 } \right).{\left( {\sqrt 3 + 1} \right)^2}\\
= \left( {2 - \sqrt 3 } \right).\left( {4 + 2\sqrt 3 } \right)\\
= 2\left( {2 - \sqrt 3 } \right).\left( {2 + \sqrt 3 } \right)\\
= 2.\left( {4 - 3} \right) = 2\\
m)\sqrt {4{x^4}\left( {{y^2} - 4y + 4} \right)} \\
= \sqrt {4{x^4}{{\left( {y - 2} \right)}^2}} \\
= 2{x^2}\left| {y - 2} \right|\\
= - 2{x^2}\left( {y - 2} \right)
\end{array}\)
