Решите неравенство: (x^4-10x^2+9)/(4x+12)<=0.

Ответ:

\[\frac{x^{4} - 10x^{2} + 9}{4x + 12} \leq 0\]

\[x^{4} - 10x^{2} + 9 =\]

\[= (x + 3)(x + 1)(x - 1)(x - 3)\]

\[Пусть\ x^{2} = y:\]

\[y^{2} - 10y + 9 = 0\]

\[D_{1} = 25 - 9 = 16\]

\[y_{1} = 5 + 4 = 9;\ \ y_{2} = 5 - 4 = 1.\]

\[1)\ x^{2} = 9\]

\[x = \pm 3.\]

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

\[x = \pm 1.\ \]

\[\frac{(x + 3)(x + 1)(x - 1)(x - 3)}{4 \cdot (x + 3)} \leq 0\]

\[x < - 3;\ - 3 < x \leq - 1;\ \ \]

\[1 \leq x \leq 3.\]