X^2+0,5x-13/x^2-0,5x-14<=1 прошууу<br>
(x²-0,5x-13)/(x²-0,5x-14)≤1 (x²-0,5x-13)/(x²-0,5x-14)-1≤0 (x²-0,5x-13-x²+0,5x+14)/(x²-0,5x-14)≤0 1/(x²-0,5x-14)≤0 ОДЗ: x²-0,5x-14≠0 |×2 2x²-x-28≠0 D=225 x₁≠4 x₂≠-3,5 1/((x-4)*(x+3,5))≤0 -∞____+_____-3,5____-_____4____+_____+∞ Ответ: x∈(-3,5;4).