a)
Vì \(\left\{ \matrix{{2^2} = 4 \hfill \cr {\left( {\sqrt 3 } \right)^2} = 3 \hfill \cr} \right.\)
mà \(4>3\) nên \(\sqrt{4} > \sqrt{3} \Leftrightarrow 2> \sqrt{3} \Leftrightarrow 2- \sqrt{3}>0 \).
\(\Leftrightarrow \left| {2 - \sqrt 3 } \right| =2- \sqrt{3}\).
Do đó: \(\sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} = \left| {2 - \sqrt 3 } \right|=2- \sqrt{3} \)
b)
Vì \(\left\{ \matrix{{3^2} = 9 \hfill \cr {\left( {\sqrt {11} } \right)^2} = 11 \hfill \cr} \right.\)
mà \( 9<11\) nên \(\sqrt{9} < \sqrt{11} \Leftrightarrow 3< \sqrt{11} \Leftrightarrow 3- \sqrt{11} <0\)
\(\Leftrightarrow \left| {3 - \sqrt {11} } \right| =-(3- \sqrt{11})=-3+\sqrt{11}\)
\(=\sqrt{11}-3\).
Do đó: \(\sqrt {{{\left( {3 - \sqrt {11} } \right)}^2}} = \left| {3 - \sqrt {11} } \right| =\sqrt{11}-3\).
c) Ta có: \(2\sqrt {{a^2}} = 2\left| a \right| = 2{\rm{a}}\) (vì \(a \ge 0\) )
d) Vì \(a < 2\) nên \(a - 2<0\).
\(\Leftrightarrow \left| a-2 \right|=-(a-2)=-a+2=2-a \)
Do đó: \(3\sqrt {{{\left( {a - 2} \right)}^2}} = 3\left| {a - 2} \right| = 3\left( {2 - a} \right) \)
\(= 6 - 3a\).
