Đáp án:
$\begin{array}{l}
a)\sqrt {{{\left( {1 - \sqrt 3 } \right)}^2}} = \left| {1 - \sqrt 3 } \right| = \sqrt 3 - 1\\
\Leftrightarrow \sqrt {{{\left( {1 - \sqrt 3 } \right)}^2}} = 1 - \sqrt 3 \left( {Sai} \right)\\
b)Dkxd:x \ge 3\\
\sqrt {x - 3} - \sqrt {x\left( {x - 3} \right)} = 0\\
\Leftrightarrow \sqrt {x - 3} - \sqrt x .\sqrt {x - 3} = 0\\
\Leftrightarrow \sqrt {x - 3} \left( {1 - \sqrt x } \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 3 = 0\\
\sqrt x = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\left( {tm} \right)\\
x = 1\left( {ktm} \right)
\end{array} \right.\\
Vay\,x = 3\\
\Leftrightarrow Đúng\\
c)\sqrt {2x - 3} \\
Dkxd:2x - 3 \ge 0\\
\Leftrightarrow 2x \ge 3\\
\Leftrightarrow x \ge \frac{3}{2}\\
Khi:x \ge - \frac{3}{2} \Leftrightarrow Sai\\
d)a - b = \left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)\\
Khi:a \ge 0;b \ge 0\\
\Leftrightarrow Sai
\end{array}$
