Canada | A Nice Algebra Problem | Math Olympiad

[@Nguyễn-j9q]

To find the value of x in this equation, you can rewrite it as: 50^(1/x) + 30^(1/x) = 18^(1/x) Now, you can see that the bases on the left side (50 and 30) can be broken down into smaller factors, while the base on the right side (18) is already smaller. Let's start breaking down the bases: 50 = 2*5^2 30 = 2*3*5 18 = 2*3^2 Rewriting the equation with these values: (2*5^2)^(1/x) + (2*3*5)^(1/x) = (2*3^2)^(1/x) 2^(1/x) * 5^(2/x) + 2^(1/x) * 3^(1/x) * 5^(1/x) = 2^(1/x) * 3^2/x Now, you can see that each term has a common factor of 2^(1/x). You can factor it out: 2^(1/x) * (5^(2/x) + 3^(1/x) * 5^(1/x)) = 3^(2/x) Now, you can simplify and solve for x: 5^(2/x) + 3^(1/x) * 5^(1/x) = 3^(2/x) 25^(1/x) + 15^(1/x) = 9^(1/x) Now, let 9 = (3^2): 25^(1/x) + 15^(1/x) = (3^2)^(1/x) 25^(1/x) + 15^(1/x) = 3 Now, let's substitute: (5^2)^(1/x) + (3*5)^(1/x) = 3 5^(2/x) + 15^(1/x) = 3 Now, we can see the equation more clearly. Since we've simplified it to this form, it's easier to solve. We're trying to solve for x in the power of the bases. Let's continue simplifying and solving for x from this point.

[@doichecabano]

Вообще-то, если степень корня не целое положииельное число, то знак корня как бы некорректен. Тогда уж степень 1/х, а не √.