Question

решите неравенство $\frac{x^{2}+x-6 }{x-9} \geq 0$

Answer 1

$\sf \displaystyle \frac{x^2+x-6}{x-9}\geq 0\\\\y=\frac{x^2+x-6}{x-9}\\y=0\\ \frac{x^2+x-6}{x-9}=0\\x^2+x-6=0\\D=b^2-4ac=1+24=25=5^2\\x_{1,2}=\frac{-1\pm 5}{2}=\left |{ {{2} \atop {-3}} \right. \\x=-3;x=2;x\neq 9\\---[-3]+++[2]---(9)+++>x\\x\in [-3;2] \cup (9;+\infty)$

Ответ: $[-3;2] \cup (9;+\infty)$