Вопрос:

Решите уравнение: 4x^4-41x^2+100=0.

Ответ:

\[4x^{4} - 41x^{2} + 100 = 0\]

\[4x^{4} - 16x^{2} - 25x^{2} + 100 = 0\]

\[4x^{2}\left( x^{2} - 4 \right) - 25\left( x^{2} - 4 \right) = 0\]

\[\left( x^{2} - 4 \right)\left( 4x^{2} - 25 \right) = 0\]

\[1)\ x^{2} = 4\]

\[x = \pm 2.\]

\[2)\ 4x^{2} = 25\]

\[x^{2} = \frac{25}{4}\]

\[x = \pm \frac{5}{2} = \pm 2,5.\]

\[Ответ:x = \pm 2;x = \pm 2,5.\]


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