Найдите область определения функции:а) f(x)=0.5- √x-3б) f(x)= √2x^2-7x+5в)...

Решите задачу:

$1)\quad f(x)=0,5-\sqrt{x-3}\\\\OOF:\; \; x-3 \geq 0\; ,\; \; x \geq 3\\\\x\in [\, 3,+\infty )\\\\2)\; \; f(x)=\sqrt{2x^2-7x+5}\\\\OOF:\; \; 2x^2-7x+5 \geq 0\; ,\\\\D=9\; ,\; \; x_1=\frac{7-3}{4}=1\; ,\; \; x_2= \frac{7+3}{4}= \frac{5}{2}=2,5\\\\+++[\, 1\, ]---[2,5]+++ \\\\x\in (-\infty ,1\, ]\cup [\, 2,5\; ;+\infty )$

$3)\; \; f(x)= \frac{x-3}{4x^2-1} \\\\OOF:\; \; 4x^2-1\ne 0\; ,\; \; x^2\ne \frac{1}{4}\; ,\; \; x\ne \pm \frac{1}{2}\\\\x\in (-\infty ,- \frac{1}{2})\cup (-\frac{1}{2},\frac{1}{2})\cup (\frac{1}{2},+\infty )\\\\4)\; \; f(x)= \frac{x}{3-x} +\frac{1}{4x^2-9} \\\\OOF:\; \; \left \{ {{3-x\ne 0} \atop {4x^2-9\ne 0}} \right. \; \; \left \{ {{x\ne 3} \atop {x^2\ne \frac{9}{4}}} \right. \; \; \left \{ {{x\ne 3} \atop {x\ne \pm 1,5}} \right. \\\\x\in (-\infty ;-1,5)\cup (-1,5\; ;\; 1,5)\cup (1,5\; ;3)\cup (3;+\infty )$