Question

2f(x)+xf(1/x)=x найдите f(10)​

Answer 1

Подставив сперва вместо x = 10, а потом x=1/10 будем иметь$\begin{cases} 2x +10y = 10\\\frac{1}{10}x+2y = \frac{1}{10}\end{cases} \leftrightarrow x=\frac{19}{3}$

Answer 2

To find the value of f(10), we first need to find the function f(x) that satisfies the given equation.

Let's substitute x = 1/10 in the given equation to get:

2f(1/10) + 10f(10) = 1/10

Now let's substitute x = 10 in the given equation to get:

2f(10) + 1/f(10) = 10

We have two equations with two variables (f(1/10) and f(10)). Let's solve for f(1/10) in the first equation:

2f(1/10) = 1/10 - 10f(10)

f(1/10) = (1/20) - 5f(10)

Now we can substitute this expression for f(1/10) into the second equation:

2f(10) + 1/[(1/20) - 5f(10)] = 10

Simplifying this equation, we get:

40f(10)^2 - 19f(10) + 2 = 0

Using the quadratic formula, we can solve for f(10):

f(10) = (19 ± sqrt(19^2 - 4402)) / (2*40)

f(10) = (19 ± sqrt(1137)) / 80

Since f(10) must be a real number, we take the positive root:

f(10) = (19 + sqrt(1137)) / 80 ≈ 0.2065

Therefore, f(10) ≈ 0.2065.