Question

Найти область определения функции y = ((22 x+1)/(√6x-√9)+((38-24)/(2x-100)

Answer 1

Y=((22x+1)/(√6x-√9))+((38-24)/(2x-100))
a)√(6x)-√9≠0, √(6x)≠√9,6x≠9, x≠3/2
b)2x-100≠0,2x≠100, x≠50
x∈(-∞,2/3)∪(2/3,50)∪(50,∞)
Ecli ect √6x=x√6, to 
a)x√6-√9≠0,x≠3/√6=3√6/√6√6=3√6/6=(√6)/2
b)x≠50
x∈(-∞,(√6)/2)∪((√6)/2,50)∪(50,∞)

Answer 2

I can help you with math problems. To find the domain of the function, you need to solve the inequalities that make the denominator and the square root expressions positive. The domain is the intersection of the solutions of these inequalities. Here are the steps:

- For the first fraction, the denominator is $\sqrt{6x}-\sqrt{9}$. This is positive when $6x > 9$ and $\sqrt{6x} > \sqrt{9}$. Solving for x, we get $x > \frac{3}{2}$ and $x > \frac{9}{6}$. The second inequality is redundant, so the solution is $x > \frac{3}{2}$. - For the second fraction, the denominator is $2x-100$. This is positive when $2x > 100$. Solving for x, we get $x > 50$. - The domain of the function is the intersection of the solutions of the two fractions. This means that x must satisfy both inequalities. The domain is $x > 50$.Ответь+подробно.+Найти+область+определения+функции+y+=+((22+x+1)/(√6x-√9)+((38-24)/(2x-100)