Question

Решите неравенство x^3+6x^2+(21x^2+3x-12)/(x-4) < =3

Answer 1

$\displaystyle x^3+6x^2+ \frac{21x^2+3x-12}{x-4} \leq 3\\\\ \frac{(x^3+6x^2)(x-4)+21x^2+3x-12}{x-4} \leq 3\\\\ \frac{x^4+6x^3-4x^3-24x^2+21x^2+3x-12}{x-4} \leq 3\\\\ \frac{x^4+2x^3-3x^2+3x-12}{x-4} \leq \frac{3(x-4)}{x-4}\\\\ \frac{x^4+2x^3-3x^2+3x-12-3x+12}{x-4} \leq 0\\\\ \frac{x^4+2x^3-3x^2}{x-4} \leq 0\\\\ \frac{x^2(x^2+2x-3)}{x-4} \leq 0\\\\ \frac{x^2(x-1)(x+3)}{(x-4)} \leq 0$

    -               +             +               -              +
----------- -3 --------- 0 ----------1---------- 4----------   

ответ x∈ (-∞; -3] ∪{0}∪[1;4)