Решите уравнение: (x^2-4)(x^2+x-2)=0.

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

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

\[x - 2 = 0\]

\[x = 2\]

\[x + 2 = 0\]

\[x = - 2\]

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

\[D = 1^{2} - 4 \cdot 1 \cdot ( - 2) = 1 + 8 =\]

\[= 9\]

\[x_{1} = \frac{- 1 + \sqrt{9}}{2} = \frac{- 1 + 3}{2} = \frac{2}{2} = 1\]

\[x_{2} = \frac{- 1 - \sqrt{9}}{2} = \frac{- 1 - 3}{2} = \frac{- 4}{2} =\]

\[= - 2\]

\[Ответ:2;\ - 2;\ 1.\]