Đáp án:
$\begin{array}{l}
a)1 = \dfrac{2}{2} = \dfrac{{\sqrt {{2^2}} }}{2} = \dfrac{{\sqrt 4 }}{2}\\
Do:\sqrt 4 > \sqrt 3 \\
\Rightarrow \dfrac{{\sqrt 4 }}{2} > \dfrac{{\sqrt 3 }}{2}\\
\Rightarrow 1 > \dfrac{{\sqrt 3 }}{2}\\
Vậy\,\dfrac{{\sqrt 3 }}{2} < 1\\
b) - 2\sqrt 5 = - \dfrac{{4\sqrt 5 }}{2} = - \dfrac{{\sqrt {16.5} }}{2} = - \dfrac{{\sqrt {80} }}{2}\\
Do:\sqrt {80} > \sqrt {10} \\
\Rightarrow - \sqrt {80} < - \sqrt {10} \\
\Rightarrow - \dfrac{{\sqrt {80} }}{2} < - \dfrac{{\sqrt {10} }}{2}\\
\Rightarrow - 2\sqrt 5 < \dfrac{{ - \sqrt {10} }}{2}\\
Vậy\, - \dfrac{{\sqrt {10} }}{2} > - 2\sqrt 5
\end{array}$
