![\frac{x-3}{x^2+x-12} \leq \frac{1}{3}](https://tex.z-dn.net/?f=+%5Cfrac%7Bx-3%7D%7Bx%5E2%2Bx-12%7D++%5Cleq++%5Cfrac%7B1%7D%7B3%7D+)
ОДЗ: x² + x - 12 ≠ 0
(x+4)(x-3)≠0
x≠-4
x≠3
![\frac{x-3}{(x-3)(x+4)} - \frac{1}{3} \leq 0\\ \frac{3x-9-x^2-x+12}{3(x-3)(x+4)} \leq 0\\ \frac{-x^2+2x+3}{(x-3)(x+4)} \leq 0\\ \frac{x^2-2x-3}{(x-3)(x+4)} \geq 0\\ \frac{(x-3)(x+1)}{(x-3)(x+4)} \geq 0](https://tex.z-dn.net/?f=%5Cfrac%7Bx-3%7D%7B%28x-3%29%28x%2B4%29%7D+-+%5Cfrac%7B1%7D%7B3%7D++%5Cleq+0%5C%5C%0A%5Cfrac%7B3x-9-x%5E2-x%2B12%7D%7B3%28x-3%29%28x%2B4%29%7D+%5Cleq+0%5C%5C%0A%5Cfrac%7B-x%5E2%2B2x%2B3%7D%7B%28x-3%29%28x%2B4%29%7D++%5Cleq+0%5C%5C%0A%5Cfrac%7Bx%5E2-2x-3%7D%7B%28x-3%29%28x%2B4%29%7D++%5Cgeq+0%5C%5C%0A%5Cfrac%7B%28x-3%29%28x%2B1%29%7D%7B%28x-3%29%28x%2B4%29%7D++%5Cgeq+0)
__+____(-4)__-___[-1]___+____(3)____+______
(-∞; -4) U [-1;3) U (3;+∞)
Наименьшее целое на отрезке (-5; 2): -1
Ответ: -1
Раскрываем скобки,приводим подобные,вычисляем
5x-y*5x+y*2a+16-
<span>2a2</span>+6a=25x2y2+8a2+16-2a=25x2y2+6a+16
B3=8;b5=32
b1*q²=8
b1*q⁴=32
q²=4
q=±2
b1=8/q²=8/4=2
2;4;8;16;32;64;128;256
s(8)=b1(qⁿ-1)/(q-1)=
2(2^8-1)/(2-1)=2*255=510