Question

Найти область определения

Answer 1

$y=\frac{\sqrt{3x^2+7x-6}}{(x+2)(x+4)}\\\\OOF:\; \left \{ {{3x^2+7x-6 \geq 0} \atop {x\ne -2,\; x\ne -4}} \right. \\\\3x^2+7x-6 \geq 0\\\\D=49-4\cdot 3\cdot (-6)=121\\\\x_1=\frac{-7-11}{6}=-3,\; x_2=\frac{-7+11}{6}=\frac{2}{3}\\\\+++(-3)---(\frac{2}{3})+++\\\\x\in (-\infty,-3]U[\frac{2}{3},+\infty)\\\\Otvet:\; x\in (-\infty,-4)U(-4,-3]U[\frac{2}{3},+\infty)$