Question

√(2х+4)/√(х^2-4) найти область определения функции

Answer 1

$\frac{\sqrt{2x+4}}{\sqrt{x^2-4}}$

ОДЗ:

$\left\{{{2x+4\geq0}\atop{x^2-4>0}}\right.$

$\left\{{{2x\geq-4}\atop{(x-2)(x+2)>0}}\right.$

$\left\{{{x\geq-2}\atop{(x-2)(x+2)>0}}\right.$

$\left\{{{x\geq-2}\atop{(x2}}\right.$

Общее решение:

$x>2$

Ответ: х ∈ (2; +∞)

Answer 2

To find the domain of the function, we need to identify the values of 'x' for which the function is defined. In this case, the function is defined as long as the denominators are non-zero, since division by zero is undefined.

The function is: $f(x) = \frac{\sqrt{2x+4}}{\sqrt{x^2-4}}$

We need to ensure that both the denominators, $\sqrt{2x+4}$ and $\sqrt{x^2-4}$, are non-zero.

1. $\sqrt{2x+4}$ is non-zero as long as $2x+4 > 0$, because the square root of a positive number is positive. So, $2x+4 > 0 \Rightarrow 2x > -4 \Rightarrow x > -2$

2. $\sqrt{x^2-4}$ is non-zero as long as $x^2 - 4 > 0$, because the square root of a positive number is positive. So, $x^2 - 4 > 0 \Rightarrow x^2 > 4 \Rightarrow x > 2$ or $x < -2$

Now, we need to find the common values that satisfy both conditions. The common region is $x > 2$ or $x < -2$. This is the domain of the function.

In interval notation, the domain of the function is $(-∞, -2) \cup (2, ∞)$.