Решите уравнение: (x+2)(x^2-4x-1)=4*|x+2|.

\[(x + 2)\left( x^{2} - 4x - 1 \right) = 4|x + 2|\]

\[(x + 2)\left( x^{2} - 4x - 1 \right) =\]

\[= - 4 \bullet (x + 2)\]

\[(x + 2)\left( x^{2} - 4x - 1 + 4 \right) = 0\]

\[(x + 2)\left( x^{2} - 4x + 3 \right) = 0\]

\[x = - 2 \Longrightarrow \ не\ подходит.\]

\[x^{2} - 4x + 3 = 0\]

\[D = ( - 4)^{2} - 4 \cdot 1 \cdot 3 =\]

\[= 16 - 12 = 4\]

\[x_{1} = \frac{4 + \sqrt{4}}{2} = \frac{4 + 2}{2} = \frac{6}{2} = 3 \Longrightarrow \ \ \]

\[\Longrightarrow \ не\ подходит.\]

\[x_{2} = \frac{4 - \sqrt{4}}{2} = \frac{4 - 2}{2} = \frac{2}{2} = 1 \Longrightarrow\]

\[\Longrightarrow \ \ не\ подходит.\]

\[(x + 2)\left( x^{2} - 4x - 1 \right) =\]

\[= 4 \bullet (x + 2)\]

\[(x + 2)\left( x^{2} - 4x - 1 - 4 \right) = 0\]

\[(x + 2)\left( x^{2} - 4x - 5 \right) = 0\]

\(x = - 2\).

\[x^{2} - 4x - 5 = 0\]

\[D = ( - 4)^{2} - 4 \cdot 1 \cdot ( - 5) =\]

\[= 16 + 20 = 36\]

\[x_{1} = \frac{4 + \sqrt{36}}{2} = \frac{4 + 6}{2} = \frac{10}{2} = 5\]

\[x_{2} = \frac{4 - \sqrt{36}}{2} = \frac{4 - 6}{2} = \frac{- 2}{2} =\]

\[= - 1\]

\[Ответ:\ - 2;\ - 1;\ \ 5.\]