Đáp án:
$\begin{array}{l}
5)\\
\overrightarrow {IA} + 2\overrightarrow {IB} + 3\overrightarrow {IC} = 0\\
\Rightarrow \overrightarrow {IA} + 2\left( {\overrightarrow {IA} + \overrightarrow {AB} } \right) + 3\left( {\overrightarrow {IA} + \overrightarrow {AC} } \right) = 0\\
\Rightarrow 6\overrightarrow {IA} + 2\overrightarrow {AB} + 3\overrightarrow {AC} = 0\\
\Rightarrow \overrightarrow {AI} = \frac{2}{6}\overrightarrow {AB} + \frac{3}{6}\overrightarrow {AC} \\
\Rightarrow \overrightarrow {AI} = \frac{1}{3}\overrightarrow {AB} + \frac{1}{2}\overrightarrow {AC} \left( {dpcm} \right)\\
6)\\
\sqrt {2x - 1} + {x^2} - 3x + 1 = 0\left( {dkxd:x \ge \frac{1}{2}} \right)\\
\Rightarrow 2\sqrt {2x - 1} + 2{x^2} - 6x + 2 = 0\\
\Rightarrow 2{x^2} - 6x + 2 = - 2\sqrt {2x - 1} \\
\Rightarrow 2{x^2} - 4x + 2 = 2x - 1 - 2\sqrt {2x - 1} + 1\\
\Rightarrow 2\left( {{x^2} - 2x + 1} \right) = {\left( {\sqrt {2x - 1} - 1} \right)^2}\\
\Rightarrow {\left( {\sqrt 2 \left( {x - 1} \right)} \right)^2} = {\left( {\sqrt {2x - 1} - 1} \right)^2}\\
\Rightarrow \left[ \begin{array}{l}
\sqrt 2 \left( {x - 1} \right) = \sqrt {2x - 1} - 1\\
\sqrt 2 \left( {x - 1} \right) = 1 - \sqrt {2x - 1}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\sqrt 2 x - \sqrt 2 = - {x^2} + 3x - 1 - 1\\
\sqrt 2 x - \sqrt 2 = 1 - \left( { - {x^2} + 3x - 1} \right)
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
{x^2} + \left( {\sqrt 2 - 3} \right)x + 2 - \sqrt 2 = 0\\
{x^2} - \left( {3 - \sqrt 2 } \right)x + \sqrt 2 + 1 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 1\left( {tm} \right)\\
x = 2 - \sqrt 2 \left( {tm} \right)\\
x = - 0,63\left( {ktm} \right)\\
x = - 3,77\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x = 1\,hoặc\,x = 2 - \sqrt 2
\end{array}$
