在 $\left(x + \frac{k}{x}\right)^5$ 的二项式展开式中，其中 $k$ 是正常数，$x^3$ 与 $x$ 的系数相同。求 $k$ 的值。

1. Write down the general term of the binomial expansion

The general term is given by:
$T_{r+1} = \binom{5}{r} x^{5-r} \left(\frac{k}{x}\right)^r$
$T_{r+1} = \binom{5}{r} k^r x^{5-2r}$

2. Find the coefficient of $x^3$

For the $x^3$ term, the power of $x$ must be 3:
$5 - 2r = 3$
$2r = 2 \implies r = 1$
The coefficient of $x^3$ is:
$\binom{5}{1} k^1 = 5k$

3. Find the coefficient of $x$

For the $x$ term, the power of $x$ must be 1:
$5 - 2r = 1$
$2r = 4 \implies r = 2$
The coefficient of $x$ is:
$\binom{5}{2} k^2 = 10k^2$

4. Equate the coefficients and solve for $k$

Since the coefficients are the same:
$10k^2 = 5k$
$10k^2 - 5k = 0$
$5k(2k - 1) = 0$
Since $k$ is a positive constant, $k \neq 0$.
$2k - 1 = 0 \implies k = \frac{1}{2}$