Đáp án:
\(\begin{array}{l}
3,\,\,\,\, - 2\sqrt 5 .\left( {\sqrt[4]{3} + \sqrt[4]{5}} \right)\\
4,\,\,\,\,M = 10\sqrt u - 13\\
5,\,\,\,\,\,A = \dfrac{{11}}{2}\sqrt x \\
6,\,\,\,\,\dfrac{{\sqrt 2 }}{2}\\
7,\,\,\,\,2\sqrt 6 \\
8,\,\,\,\,\dfrac{{5x - 4y + 1}}{{\sqrt {xy} }}\\
9,\,\,\,\, - 2\\
10,\,\,\,\,P = 2
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
3,\\
2\sqrt {40\sqrt {12} } - 2\sqrt {\sqrt {75} } - 3\sqrt {20\sqrt {48} } \\
= 2.\sqrt {40.\sqrt {4.3} } - 2\sqrt {\sqrt {25.5} } - 3\sqrt {20.\sqrt {16.3} } \\
= 2\sqrt {40.\sqrt {{2^2}.3} } - 2\sqrt {\sqrt {{5^2}.5} } - 3\sqrt {20.\sqrt {{4^2}.3} } \\
= 2\sqrt {40.2\sqrt 3 } - 2\sqrt {5.\sqrt 5 } - 3\sqrt {20.4\sqrt 3 } \\
= 2\sqrt {80\sqrt 3 } - 2\sqrt {5\sqrt 5 } - 3\sqrt {80\sqrt 3 } \\
= - \sqrt {80\sqrt 3 } - 2\sqrt {5\sqrt 5 } \\
= - \sqrt {{4^2}.5\sqrt 3 } - 2\sqrt {5\sqrt 5 } \\
= - 2\sqrt {5\sqrt 3 } - 2\sqrt {5\sqrt 5 } \\
= - 2\sqrt 5 .\sqrt {\sqrt 3 } - 2.\sqrt 5 .\sqrt {\sqrt 5 } \\
= - 2\sqrt 5 .\left( {\sqrt {\sqrt 3 } + \sqrt {\sqrt 5 } } \right)\\
= - 2\sqrt 5 .\left( {\sqrt[4]{3} + \sqrt[4]{5}} \right)\\
4,\\
u > 0 \Rightarrow \left| u \right| = u\\
M = 4\sqrt {25u} - \dfrac{{15}}{2}\sqrt {\dfrac{{16u}}{9}} - \dfrac{2}{u}\sqrt {\dfrac{{169{u^2}}}{4}} \\
= 4\sqrt {{5^2}.u} - \dfrac{{15}}{2}.\sqrt {{{\left( {\dfrac{4}{3}} \right)}^2}.u} - \dfrac{2}{u}.\sqrt {{{\left( {\dfrac{{13u}}{2}} \right)}^2}} \\
= 4.5\sqrt u - \dfrac{{15}}{2}.\dfrac{4}{3}\sqrt u - \dfrac{2}{u}.\dfrac{{13u}}{2}\\
= 20\sqrt u - 10\sqrt u - 13\\
= 10\sqrt u - 13\\
5,\\
x \ge 0 \Rightarrow \left| x \right| = x\\
A = 4\sqrt {\dfrac{{25x}}{4}} - \dfrac{8}{3}\sqrt {\dfrac{{9x}}{4}} - \dfrac{4}{{3x}}.\sqrt {\dfrac{{9{x^3}}}{{64}}} \\
= 4\sqrt {\dfrac{{25}}{4}.x} - \dfrac{8}{3}.\sqrt {\dfrac{9}{4}.x} - \dfrac{4}{{3x}}.\sqrt {\dfrac{{9{x^2}}}{{64}}.x} \\
= 4.\sqrt {{{\left( {\dfrac{5}{2}} \right)}^2}.x} - \dfrac{8}{3}\sqrt {{{\left( {\dfrac{3}{2}} \right)}^2}.x} - \dfrac{4}{{3x}}\sqrt {{{\left( {\dfrac{{3x}}{8}} \right)}^2}.x} \\
= 4.\dfrac{5}{2}\sqrt x - \dfrac{8}{3}.\dfrac{3}{2}.\sqrt x - \dfrac{4}{{3x}}.\dfrac{{3x}}{8}.\sqrt x \\
= 10\sqrt x - 4\sqrt x - \dfrac{1}{2}\sqrt x \\
= \dfrac{{11}}{2}\sqrt x \\
6,\\
3\sqrt {\dfrac{1}{2}} + \sqrt {4,5} - \sqrt {12,5} \\
= 3\sqrt {\dfrac{1}{2}} + \sqrt {\dfrac{9}{2}} - \sqrt {\dfrac{{25}}{2}} \\
= 3\sqrt {\dfrac{1}{2}} + \sqrt {{3^2}.\dfrac{1}{2}} - \sqrt {{5^2}.\dfrac{1}{2}} \\
= 3\sqrt {\dfrac{1}{2}} + 3\sqrt {\dfrac{1}{2}} - 5\sqrt {\dfrac{1}{2}} \\
= \sqrt {\dfrac{1}{2}} = \dfrac{1}{{\sqrt 2 }} = \dfrac{{\sqrt 2 }}{2}\\
7,\\
42\sqrt {\dfrac{{25}}{6}} - 10\sqrt {\dfrac{3}{2}} - 12\sqrt {\dfrac{{98}}{3}} \\
= 42\sqrt {25.\dfrac{1}{6}} - 10\sqrt {9.\dfrac{1}{6}} - 12\sqrt {196.\dfrac{1}{6}} \\
= 42\sqrt {{5^2}.\dfrac{1}{6}} - 10\sqrt {{3^2}.\dfrac{1}{6}} - 12\sqrt {{{14}^2}.\dfrac{1}{6}} \\
= 42.5\sqrt {\dfrac{1}{6}} - 10.3\sqrt {\dfrac{1}{6}} - 12.14\sqrt {\dfrac{1}{6}} \\
= 210\sqrt {\dfrac{1}{6}} - 30\sqrt {\dfrac{1}{6}} - 168\sqrt {\dfrac{1}{6}} \\
= 12\sqrt {\dfrac{1}{6}} \\
= \dfrac{{12}}{{\sqrt 6 }}\\
= \dfrac{{12\sqrt 6 }}{6}\\
= 2\sqrt 6 \\
8,\\
5\sqrt {\dfrac{x}{y}} - 4\sqrt {\dfrac{y}{x}} + \sqrt {\dfrac{1}{{xy}}} \\
= 5\sqrt {{x^2}.\dfrac{1}{{xy}}} - 4\sqrt {{y^2}.\dfrac{1}{{xy}}} + \dfrac{1}{{\sqrt {xy} }}\\
= 5\left| x \right|.\sqrt {\dfrac{1}{{xy}}} - 4.\left| y \right|.\sqrt {\dfrac{1}{{xy}}} + \dfrac{1}{{\sqrt {xy} }}\\
= 5x.\dfrac{1}{{\sqrt {xy} }} - 4y.\dfrac{1}{{\sqrt {xy} }} + \dfrac{1}{{\sqrt {xy} }}\\
= \dfrac{{5x - 4y + 1}}{{\sqrt {xy} }}\\
9,\\
\dfrac{3}{{\sqrt 5 + \sqrt 2 }} - \dfrac{4}{{3 - \sqrt 5 }} + \dfrac{1}{{\sqrt 2 - 1}}\\
= \dfrac{{3.\left( {\sqrt 5 - \sqrt 2 } \right)}}{{\left( {\sqrt 5 - \sqrt 2 } \right).\left( {\sqrt 5 + \sqrt 2 } \right)}} - \dfrac{{4.\left( {3 + \sqrt 5 } \right)}}{{\left( {3 - \sqrt 5 } \right).\left( {3 + \sqrt 5 } \right)}} + \dfrac{{\sqrt 2 + 1}}{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 + 1} \right)}}\\
= \dfrac{{3.\left( {\sqrt 5 - \sqrt 2 } \right)}}{{{{\sqrt 5 }^2} - {{\sqrt 2 }^2}}} - \dfrac{{4.\left( {3 + \sqrt 5 } \right)}}{{{3^2} - {{\sqrt 5 }^2}}} + \dfrac{{\sqrt 2 + 1}}{{{{\sqrt 2 }^2} - {1^2}}}\\
= \dfrac{{3.\left( {\sqrt 5 - \sqrt 2 } \right)}}{{5 - 2}} - \dfrac{{4.\left( {3 + \sqrt 5 } \right)}}{{9 - 5}} + \dfrac{{\sqrt 2 + 1}}{{2 - 1}}\\
= \dfrac{{3.\left( {\sqrt 5 - \sqrt 2 } \right)}}{3} - \dfrac{{4.\left( {3 + \sqrt 5 } \right)}}{4} + \dfrac{{\sqrt 2 + 1}}{1}\\
= \left( {\sqrt 5 - \sqrt 2 } \right) - \left( {3 + \sqrt 5 } \right) + \left( {\sqrt 2 + 1} \right)\\
= \sqrt 5 - \sqrt 2 - 3 - \sqrt 5 + \sqrt 2 + 1\\
= - 2\\
10,\\
P = \dfrac{{3 + 2\sqrt 3 }}{{\sqrt 3 }} + \dfrac{{2 + \sqrt 2 }}{{\sqrt 2 + 1}} - \left( {\sqrt 2 + \sqrt 3 } \right)\\
= \dfrac{{{{\sqrt 3 }^2} + 2\sqrt 3 }}{{\sqrt 3 }} + \dfrac{{{{\sqrt 2 }^2} + \sqrt 2 }}{{\sqrt 2 + 1}} - \left( {\sqrt 2 + \sqrt 3 } \right)\\
= \dfrac{{\sqrt 3 .\left( {\sqrt 3 + 2} \right)}}{{\sqrt 3 }} + \dfrac{{\sqrt 2 .\left( {\sqrt 2 + 1} \right)}}{{\sqrt 2 + 1}} - \left( {\sqrt 2 + \sqrt 3 } \right)\\
= \sqrt 3 + 2 + \sqrt 2 - \sqrt 2 - \sqrt 3 \\
= 2
\end{array}\)
