Помогитееее!Пусть f(x) = 2 - корень(1-x) ; g(x) = 1+2x / 3+xНайдите область определения...

Решите задачу:

$f(x) = 2 - \sqrt{1-x} \\\ g(x) = \frac{1+2x}{3+x}$

$f(x) +g(x) = 2 - \sqrt{1-x}+ \frac{1+2x}{3+x} \\\ \left \{ {{1-x \geq 0} \atop {3+x \neq 0}} \right. \\\ \left \{ {{x \leq 1} \atop {x \neq -3}} \right. \\\ x\in(-\infty; -3)\cup(-3;1]$

$f(x) = 2 - \sqrt{1-x} \\\ g(x) = \frac{1+2x}{3+x} \\\ \frac{f(x)}{g(x)} = \frac{(2 - \sqrt{1-x})(3+x)}{1+2x} \\\ \left \{ {{1-x \geq 0} \atop {1+2x \neq 0}} \right. \\\ \left \{ {{x \leq 1} \atop {x \neq -0.5}} \right. \\\ x\in(-\infty; -0.5)\cup(-0.5;1]$