Решите неравенство (x^2-5)(4x^2-x-5)<(x^2-3)(4x^2-x-5).

Ответ:

\[\left( x^{2} - 5 \right)\left( 4x^{2} - x - 5 \right) < (x^{2} - 3)(4x^{2} - x - 5)\]

\[Упростим\ левую\ часть:\]

\[4x^{4} + x^{3} - 5x^{2} - 20x^{2} + 5x + 25 =\]

\[= 4x^{4} - x^{3} - 25x^{2} + 5x + 25\]

\[Упростим\ правую\ часть:\]

\[4x^{4} - x^{3} - 5x^{2} - 12x^{2} + 3x + 15 =\]

\[= 4x^{4} - x^{3} - 17x^{2} + 3x + 15\]

\[4x^{4} - x^{3} - 25x^{2} + 5x + 25 -\]

\[- \left( 4x^{4} - x^{3} - 17x^{2} + 3x + 15 \right) < 0\]

\[4x^{4} - x^{3} - 25x^{2} + 5x + 25 - 4x^{4} + x^{3} +\]

\[+ 17x^{2} - 3x - 15 < 0\]

\[- 8x^{2} + 2x + 10 < 0\]

\[D = 4 + 320 = 324\]

\[x_{1} = \frac{- 2 + 18}{- 16} = - 1;\ \ x_{2} = \frac{- 2 - 18}{- 16} = 1\frac{1}{4}\]

\[(x + 1)(x - 1,25) < 0\]

\[Ответ:x \in ( - \infty; - 1) \cup (1,25; + \infty).\]