Question

Найдите область определения функции y=3/x2-1 и y=2/x - 1/x-5 Срочнаааа, аааа, плес.

Answer 1

У=3/(х²-1)
х²-1≠0
(х-1)(х+1)≠0
х≠1; х≠-1
Ответ: х- любое, кроме х=1; х= -1
или х∈(-∞; -1)∪(-1;1)∪(1;+∞)
у=2/х- 1/(х-5)
х≠0;  х-5≠0
х≠0; х≠5
Ответ: х-любое, кроме х=0; х=5
или х∈(-∞;0)∪(0;5)∪(5;+∞)

Answer 2

Finding the Domain of the Functions

To find the domain of the given functions, we need to determine the values of \( x \) for which the functions are defined. Let's start by analyzing each function separately.

Function 1: \( y = \frac{3}{{x^2 - 1}} \)

The domain of a rational function like \( y = \frac{3}{{x^2 - 1}} \) is all real numbers except the values of \( x \) that make the denominator equal to zero. In this case, the denominator \( x^2 - 1 \) cannot be zero.

The expression \( x^2 - 1 \) can be factored as \( (x + 1)(x - 1) \). Therefore, the domain of the function is all real numbers except \( x = 1 \) and \( x = -1 \).

Domain of the function 1: The domain of the function \( y = \frac{3}{{x^2 - 1}} \) is all real numbers except \( x = 1 \) and \( x = -1\).

Function 2: \( y = \frac{2}{x} - \frac{1}{{x - 5}} \)

For the function \( y = \frac{2}{x} - \frac{1}{{x - 5}} \), we need to consider the values of \( x \) that make the denominators \( x \) and \( x - 5 \) equal to zero.

The first term \( \frac{2}{x} \) is defined for all real numbers except \( x = 0 \). The second term \( \frac{1}{{x - 5}} \) is defined for all real numbers except \( x = 5 \).

Therefore, the domain of the function \( y = \frac{2}{x} - \frac{1}{{x - 5}} \) is all real numbers except \( x = 0 \) and \( x = 5 \).

Domain of the function 2: The domain of the function \( y = \frac{2}{x} - \frac{1}{{x - 5}} \) is all real numbers except \( x = 0 \) and \( x = 5 \).

I hope this helps! If you have any further questions or need additional assistance, feel free to ask.