\(\begin{array}{l}
a)\,\sqrt {{x^2} + 1} + \sqrt {{{\left( {2x - 1} \right)}^2} + 4} = 0\\
Vi\,\,\left\{ \begin{array}{l}
\sqrt {{x^2} + 1} \ge 1\\
\sqrt {{{\left( {2x - 1} \right)}^2} + 4} \ge 4
\end{array} \right.\\
\Rightarrow \sqrt {{x^2} + 1} + \sqrt {{{\left( {2x - 1} \right)}^2} + 4} \ge 5 > 0\\
\Rightarrow PTVN
\end{array}\)
\(\begin{array}{l}
b)\,pt \Leftrightarrow \sqrt {2{{\left( {x - 1} \right)}^2} + 1} + \sqrt {3{{\left( {x - 1} \right)}^2} + 4} = 3 - {\left( {x - 1} \right)^2}\left( * \right)\\
Ta\,co:\,\,\sqrt {2{{\left( {x - 1} \right)}^2} + 1} + \sqrt {3{{\left( {x - 1} \right)}^2} + 4} \ge 1 + 2 = 3\\
3 - {\left( {x - 1} \right)^2} \le 3\\
\Rightarrow \left( * \right)\,xay\,ra\,khi\,x - 1 = 0 \Leftrightarrow x = 1
\end{array}\)
