$\frac{2}{\sqrt{x+1}+\sqrt{3-x}}$ = 1 + $\sqrt{3+2x-x^{2}}$
$\begin{array}{l}
dk: - 1 < x < 3\\
\frac{2}{{\sqrt {x + 1`} + \sqrt {3 - x} }} = 1 + \sqrt {3 + 2x - {x^2}} \\
dat:\left\{ \begin{array}{l}
\sqrt {x + 1} = a > 0\\
\sqrt {3 - x} = b > 0
\end{array} \right. \to \left\{ \begin{array}{l}
\sqrt {3 + 2x - {x^2}} = ab\\
{a^2} + {b^2} = 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\frac{2}{{a + b}} = 1 + ab \leftrightarrow 2 = a + b + ab\left( {a + b} \right)\\
{a^2} + {b^2} = 4 \leftrightarrow {\left( {a + b} \right)^2} - 2ab = 4
\end{array} \right.\\
dat:\left\{ \begin{array}{l}
a + b = s\\
ab = p
\end{array} \right. \to \left\{ \begin{array}{l}
s + sp = 2\\
{s^2} - 2p = 4
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
p = \frac{{{s^2} - 4}}{2}\\
s + \frac{{{s^3} - 4s}}{2} = 2 \leftrightarrow {s^3} - 2s - 4 = 0 \leftrightarrow \left( {s - 2} \right)\left( {{s^2} + 2s + 2} \right) = 0
\end{array} \right.
\end{array}$
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