x\\x\in [-3;2] \cup (9;+\infty)" alt="\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)" align="absmiddle" class="latex-formula">
Ответ: ![[-3;2] \cup (9;+\infty)](https://tex.z-dn.net/?f=%5B-3%3B2%5D%20%5Ccup%20%289%3B%2B%5Cinfty%29 "[-3;2] \cup (9;+\infty)")