$\begin{array}{l} 1)\sqrt {x + 4}  = 2 - x\\ Dk:x \ge  - 4\\  \Leftrightarrow \left\{ \begin{array}{l} 2 - x \ge 0\\ x + 4 = {(2 - x)^2} \end{array} \right.\\  \Leftrightarrow \left\{ \begin{array}{l} x \le 2\\ x + 4 = 3 - 4x + {x^2} \end{array} \right.\\  \Leftrightarrow \left\{ \begin{array}{l} x \le 2\\ {x^2} - 5x - 1 = 0 \end{array} \right.\\  \Leftrightarrow \left\{ {\begin{array}{*{20}{c}} {x \le 2}\\ {\left[ \begin{array}{l} x = \frac{{5 + \sqrt {21} }}{2}\\ x = \frac{{5 - \sqrt {21} }}{2} \end{array} \right.} \end{array}} \right.\\  \Rightarrow x = \frac{{5 - \sqrt {21} }}{2}\\ 2)\frac{{ - x + 12}}{{{x^2} + 2x}} - \frac{3}{x} = \frac{{x + 5}}{{x + 2}}\\ dk:x \ne 0;x \ne  - 2\\  \Leftrightarrow \frac{{ - x + 12}}{{x(x + 2)}} - \frac{3}{x} = \frac{{x - 5}}{{x + 2}}\\  \Leftrightarrow \frac{{ - x + 12}}{{x(x + 2)}} - \frac{{3(x + 2)}}{{x(x + 2)}} = \frac{{x(x - 5)}}{{x(x + 2)}}\\  \Leftrightarrow  - x + 12 - (3x + 6) = {x^2} - 5x\\  \Leftrightarrow  - x + 12 - 3x - 6 = {x^2} - 5x\\  \Leftrightarrow {x^2} - x - 6 = 0\\  \Leftrightarrow \left[ \begin{array}{l} x = 3(tm)\\ x =  - 2(l) \end{array} \right.\\  \Rightarrow x = 3 \end{array}$