Question

найти область определения функции.​

Answer 1

Ответ:

======================================$y=\sqrt{5x^2-3x-2}+\frac{1}{2x+3}\\\\Dy: \left \{ {{5x^2-3x-2\geq 0} \atop {2x+3\neq 0}} \right.\\\\x_{1,2}=\frac{3^+_-\sqrt{9+40} }{10}=\frac{3^+_-7}{10}\\x_1=1\\x_2=-\frac{2}{5}\\\\\left \{ {{5(x-1)(x+\frac{2}{5})\geq 0 } \atop {x\neq -\frac{3}{2} }} \right. \\\\+++(-\frac{3}{2})++[-\frac{2}{5} ]-----[1]+++++$

x∈(-∞;-3/2)∪(-3/2;-2/5]∪[1;+∞)