Решите неравенство: (x^4-17x^2+16)/(5x+20)<=0.

Ответ:

\[\frac{x^{4} - 17x^{2} + 16}{5x + 20} \leq 0\]

\[x^{4} - 17x^{2} + 16 =\]

\[= (x + 4)(x + 1)(x - 1)(x - 4)\]

\[Пусть\ x^{2} = t \geq 0:\]

\[t^{2} - 17t + 16 = 0\]

\[t_{1} + t_{2} = 17;\ \ t_{1} \cdot t_{2} = 16\]

\[t_{1} = 1;\ \ t_{2} = 16.\]

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

\[x = \pm 1.\]

\[2)\ x^{2} = 16\]

\[x = \pm 4.\ \]

\[x < - 4;\ - 4 < x \leq - 1;\ \ \]

\[1 \leq x \leq 4.\]