Решите неравенство a) 3x²-12/1-11x >0 δ) 2x²-5x+2/x+4 <0

а) \[\frac{3x^2 - 12}{1 - 11x} > 0\] \( \frac{3(x^2 - 4)}{1 - 11x} > 0 \) \( \frac{3(x - 2)(x + 2)}{1 - 11x} > 0 \)

        - - - (-2) + + + (2) - - - (1/11) + + +
        ---------|---------|---------|---------> x
                 -2       1/11       2
    

\(x \in (-\infty; -2] \cup (\frac{1}{11}; 2])\)

б) \[\frac{2x^2 - 5x + 2}{x + 4} < 0\] \[2x^2 - 5x + 2 = 0\] \[D = (-5)^2 - 4 \cdot 2 \cdot 2 = 25 - 16 = 9\] \[x_1 = \frac{5 + \sqrt{9}}{2 \cdot 2} = \frac{5 + 3}{4} = \frac{8}{4} = 2\] \[x_2 = \frac{5 - \sqrt{9}}{2 \cdot 2} = \frac{5 - 3}{4} = \frac{2}{4} = \frac{1}{2} = 0.5\] \[\frac{2(x - 2)(x - 0.5)}{x + 4} < 0\]

        - - - (-4) + + + (0.5) - - - (2) + + + 
        ---------|---------|---------|---------> x
                -4        0.5        2
    

\(x \in (-4; 0.5) \cup (2; +\infty)\)