\(\begin{array}{l}
{x^2} - 6x + 13 = \sqrt {x + 1} + \sqrt {7 - x} \\
DK:\, - 1 \le x \le 7\\
Dat\,\sqrt {x + 1} + \sqrt {7 - x} = t\,\,\left( {t \ge 0} \right)\\
\Rightarrow {t^2} = x + 1 + 2\sqrt {\left( {x + 1} \right)\left( {7 - x} \right)} + 7 - x\\
\Leftrightarrow {t^2} = 8 + 2\sqrt { - {x^2} + 6x + 7} \\
\Leftrightarrow - {x^2} + 6x + 7 = {\left( {\dfrac{{{t^2} - 8}}{2}} \right)^2}\\
\Leftrightarrow {x^2} - 6x + 13 = - {\left( {\dfrac{{{t^2} - 8}}{2}} \right)^2} + 20\\
pt \Rightarrow 20 - {\left( {\dfrac{{{t^2} - 8}}{2}} \right)^2} = t\\
\Leftrightarrow {t^4} - 16{t^2} + 64 - 80 + 4t = 0\\
\Leftrightarrow {t^4} - 16{t^2} + 4t - 16 = 0\\
\Leftrightarrow {t^2}\left( {t - 4} \right)\left( {t + 4} \right) + 4\left( {t - 4} \right) = 0\\
\Leftrightarrow \left( {t - 4} \right)\left( {{t^3} + 4{t^2} + 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
t = 4\\
{t^3} + 4{t^2} + 4 = 0\left( {VN\,\,do\,\,t \ge 0} \right)
\end{array} \right.\\
\Rightarrow {x^2} - 6x + 13 = - {\left( {\frac{{{4^2} - 8}}{2}} \right)^2} + 20\\
\Leftrightarrow {x^2} - 6x + 13 = 4\\
\Leftrightarrow {\left( {x - 3} \right)^2} = 0\\
\Leftrightarrow x = 3\left( {tm} \right)
\end{array}\)
