Đáp án:
\(\begin{array}{l}
7)\dfrac{{\sqrt 2 }}{{\sqrt {10} - 3}}\\
8)\dfrac{{\sqrt {15} - \sqrt 5 }}{2}\\
9)2\\
10)2\\
11) - 2\left( {x - 1} \right)\\
12) - xy\\
13) - \left( {y - 1} \right)\\
14) = 1
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
7)\sqrt {\dfrac{2}{{{{\left( {3 - \sqrt {10} } \right)}^2}}}} \\
= \dfrac{{\sqrt 2 }}{{\left| {3 - \sqrt {10} } \right|}}\\
= \dfrac{{\sqrt 2 }}{{\sqrt {10} - 3}}\left( {do:\sqrt {10} > 3} \right)\\
8)\sqrt {\dfrac{{5{{\left( {1 - \sqrt 3 } \right)}^2}}}{4}} \\
= \dfrac{{\left| {1 - \sqrt 3 } \right|\sqrt 5 }}{2} = \dfrac{{\left( {\sqrt 3 - 1} \right)\sqrt 5 }}{2}\\
= \dfrac{{\sqrt {15} - \sqrt 5 }}{2}\\
9)x{y^2}.\sqrt {\dfrac{4}{{{x^2}{y^4}}}} \\
= x{y^2}.\dfrac{2}{{\left| {x{y^2}} \right|}} = x{y^2}.\dfrac{2}{{x{y^2}}} = 2\\
10)\left( {x - 1} \right)\sqrt {\dfrac{4}{{{{\left( {x - 1} \right)}^2}}}} = \left( {x - 1} \right).\dfrac{2}{{\left| {x - 1} \right|}}\\
= \left( {x - 1} \right).\dfrac{2}{{x - 1}} = 2\\
11){\left( {x - 1} \right)^2}\sqrt {\dfrac{4}{{{{\left( {x - 1} \right)}^2}}}} = {\left( {x - 1} \right)^2}.\dfrac{2}{{\left| {x - 1} \right|}}\\
= - {\left( {x - 1} \right)^2}.\dfrac{2}{{x - 1}} = - 2\left( {x - 1} \right)\\
12)\dfrac{{{x^2}}}{y}.\sqrt {\dfrac{{{y^4}}}{{{x^2}}}} \\
= \dfrac{{{x^2}}}{y}.\dfrac{{{y^2}}}{{\left| x \right|}} = - \dfrac{{{x^2}}}{y}.\dfrac{{{y^2}}}{x}\\
= - xy\\
13)\dfrac{{x - 2}}{{y + 1}}.\sqrt {\dfrac{{{{\left( {y + 1} \right)}^4}}}{{{{\left( {x - 2} \right)}^2}}}} \\
= \dfrac{{x - 2}}{{y + 1}}.\dfrac{{{{\left( {y + 1} \right)}^2}}}{{\left| {x - 2} \right|}}\\
= - \dfrac{{x - 2}}{{y + 1}}.\dfrac{{{{\left( {y + 1} \right)}^2}}}{{x - 2}}\\
= - \left( {y - 1} \right)\\
14)\dfrac{{{{\left( {x + 2} \right)}^2}}}{{y - 2}}.\sqrt {\dfrac{{{{\left( {y - 2} \right)}^2}}}{{{{\left( {x + 2} \right)}^4}}}} \\
= \dfrac{{{{\left( {x + 2} \right)}^2}}}{{y - 2}}.\dfrac{{\left| {y - 2} \right|}}{{{{\left( {x + 2} \right)}^2}}}\\
= \dfrac{{{{\left( {x + 2} \right)}^2}}}{{y - 2}}.\dfrac{{y - 2}}{{{{\left( {x + 2} \right)}^2}}} = 1
\end{array}\)
