Помогитееее!Пусть f(x) = 2 - корень(1-x) ; g(x) = 1+2x / 3+xНайдите область определения функции:а)y=f(x)+g(x)б)y=f(x) / g(x)

[latex]f(x) = 2 - \sqrt{1-x} \\\ g(x) = \frac{1+2x}{3+x} [/latex] [latex]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][/latex] [latex]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][/latex]

a) [latex]y=2- \sqrt{1-x}+ \frac{1+2x}{3+x} [/latex] [latex]1-x \geq 0[/latex] [latex]x \leq 1[/latex] [latex]3+x \neq 0[/latex] [latex]x \neq -3[/latex] D(y)=(-∞;-3)U(-3;1] 2) [latex]y= \frac{2- \sqrt{1-x}*(3+x)}{1+2x} [/latex] [latex]1-x \geq 0[/latex] [latex]x \leq 1[/latex] [latex]1+2x \neq 0[/latex] [latex]x \neq -\frac{1}{2}[/latex] D(y)= (-∞;-0,5)U(-0,5;1]