Стороны этого квадрата будут 4 см,потому-что ПЕРИМЕТР=2+2+6+6=16;16:4=4
Числа х, (х+1), (х+2), (х+3).
За умовою задачi
(x+2)(x+3)-x(x+1)=34
x^2+3x+2x+6-x^2-x=34
4x=34
x=34/4
x=8,5
int (4√x) dx = 4* x^(1/2 + 1) / 1/2 + C = 4 * x^(3/2) * 2 + C = 8*x^(3/2) + C
![\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