Решите неравенство ( х²+x-45)÷(x-6)≤(3x+1)÷2

$\frac{x^{2}+x-45}{x-6}\leq \frac{3x+1}{2}$

$\frac{x^{2}+x-45}{x-6} -\frac{3x+1}{2}\leq 0$

$\frac{(x^{2}+x-45)*2-(3x+1)*(x-6)}{(x-6)*2}\leq 0$

$\frac{2x^{2}+2x-90-3x^{2}-x+18x+6}{(x-6)*2}\leq 0$

$\frac{-x^{2}+19x-84}{(x-6)*2}\leq 0$

метод интервалов:

1. $\frac{-x^{2}+19x-84}{(x-6)*2}=0$

$\left \{ {{-x^{2}+19x-84=0} \atop {(x-6)*2\neq}} \right. 0$

-x²+19x-84=0. D=19²-4*(-1)*(-84)=361-336=25

x₁=7, x₂=12

x-6≠0. x≠6

++++++(6)---------[7]+++++++[12]---------->x

3. x>6, x≤7, x≥12

ответ: х принадлежит (6;7] u [12;∞)