Question

Помогите!

Найти область определения функции

1)f(x)=x/(x³+8)(x-3)²

2)f(x)=1/x-3

3)f(x)=x²-1

4)f(x)=|x|-2

5)f(x)=2x/x²-5x+4

Answer 1

$1)\; \; f(x)=\frac{x}{(x^3+8)(x-3)^2}\; \; \Rightarrow \; \; (x^3+8)(x-3)^2\ne 0\\\\\left \{ {{x^3+8\ne 0} \atop {x-3\ne 0}} \right. \; \left \{ {{(x+2)(x^2-2x+4)\ne 0} \atop {x\ne 3}} \right. \; \left \{ {{x\ne -2\; \; (x^2-2x+4>0)} \atop {x\ne 3\qquad \qquad \qquad }} \right. \\\\\underline {x\in (-\infty ,-2)\cup (-2,3)\cup (3,+\infty )}\\\\2)\; \; f(x)=\frac{1}{x-3}\; \; \Rightarrow \; \; x-3\ne 0\; ,\; x\ne 3\\\\\underline {x\in (-\infty ,3)\cup (3,+\infty )}\\\\3)\; \; f(x)=x^2-1\; \; \Rightarrow \; \; \underline {x\in (-\infty ,+\infty )}$

$4)\; \; f(x)=|x|-2\; \; \Rightarrow \; \; \underline {x\in (-\infty ,+\infty )}\\\\5)\; \; f(x)=\frac{2x}{x^2-5x+4}\; \; \Rightarrow \; \; x^2-5x+4\ne 0\\\\x^2-5x+4=0\; \; \to \; \; x_1=1\; ,\; \; x_2=4\; \; (teorema\; Vieta)\\\\x\ne 1\; ,\; \underline {x\ne 4\\\\x\in (-\infty ,1)\cup (1,4)\cup (4,+\infty )}$