Đáp án:
\(\begin{array}{l}
g,\,\,\,2\sqrt {31} > 10\\
h,\,\,\,\sqrt 3 > - 12\\
i,\,\,\, - 5 > - \sqrt {29} \\
j,\,\,\,2\sqrt 5 > \sqrt {19} \\
k,\,\,\,\sqrt {\sqrt 3 } < \sqrt 2 \\
l,\,\,\,\sqrt {2\sqrt 3 } < \sqrt {3\sqrt 2 } \\
m,\,\,\,2 + \sqrt 6 < 5\\
n,\,\,\,7 - 2\sqrt 2 > 4\\
o,\,\,\,\sqrt {15} + \sqrt 8 < 7
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
g,\\
2\sqrt {31} > 2\sqrt {25} = 2.5 = 10\\
\Rightarrow 2\sqrt {31} > 10\\
h,\\
\sqrt 3 > 0 > - 12\\
\Rightarrow \sqrt 3 > - 12\\
i,\\
5 = \sqrt {25} < \sqrt {29} \\
\Rightarrow - 5 > - \sqrt {29} \\
j,\\
2\sqrt 5 = \sqrt {{2^2}} .\sqrt 5 = \sqrt 4 .\sqrt 5 = \sqrt {20} > \sqrt {19} \\
\Rightarrow 2\sqrt 5 > \sqrt {19} \\
k,\\
\sqrt {\sqrt 3 } < \sqrt {\sqrt 4 } = \sqrt {\sqrt {{2^2}} } = \sqrt 2 \\
\Rightarrow \sqrt {\sqrt 3 } < \sqrt 2 \\
l,\\
2\sqrt 3 = \sqrt {{2^2}.3} = \sqrt {12} < \sqrt {18} = \sqrt {{3^2}.2} = 3\sqrt 2 \\
\Rightarrow \sqrt {2\sqrt 3 } < \sqrt {3\sqrt 2 } \\
m,\\
2 + \sqrt 6 < 2 + \sqrt 9 = 2 + 3 = 5\\
\Rightarrow 2 + \sqrt 6 < 5\\
n,\\
2\sqrt 2 = \sqrt {{2^2}.2} = \sqrt 8 < \sqrt 9 = 3\\
\Rightarrow 7 - 2\sqrt 2 > 7 - 3\\
\Leftrightarrow 7 - 2\sqrt 2 > 4\\
o,\\
\sqrt {15} + \sqrt 8 < \sqrt {16} + \sqrt 9 = 4 + 3 = 7\\
\Rightarrow \sqrt {15} + \sqrt 8 < 7
\end{array}\)
