[tex] \frac{2x + 3}{ { \times }^{2} + x - 12 } \leqslant \frac{1}{2} [/tex]
метод интервалов решите плз
метод интервалов решите плз
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(4x+6-x²-x+12)/(x²+x-12)≤0
(-x²+3x+18)/(x²+x-12)≤0
(x²-3x-18)/(x²+x-12)≤0
(x-6)(x+3)/(x+4)(x-3)≤0
x=6 x=-3 x=-4 x=3
+ _ + _ +
------(-4)--------[-3]---------(3)--------[6]-------
x∈(-4;-3] U (3;6]