Решить уравнение: (x^2-5)^2+4(x^2-5)+3=0.

Ответ:

\[\left( x^{2} - 5 \right)^{2} + 4 \cdot \left( x^{2} - 5 \right) + 3 = 0\]

\[x^{2} - 5 = t:\]

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

\[D_{1} = 4 - 3 = 1\]

\[t_{1} = - 2 + 1 = - 1;\]

\[t_{2} = - 2 - 1 = - 3.\]

\[1)\ x^{2} - 5 = - 1\]

\[x^{2} = 4\]

\[x = \pm 2.\]

\[2)\ x^{2} - 5 = - 3\]

\[x^{2} = 2\]

\[x = \pm \sqrt{2}.\]

\[Ответ:x = \pm 2;x = \pm \sqrt{2}.\]