38. Define the operations: $a \oplus b=\frac{1}{a}+\frac{1}{b}, a \quad b=\frac{1}{a}-\frac{1}{b}$, according to this definition, the solution to the equation $\frac{x(x+1)}{x \oplus(x+1)}=\frac{1}{3}$ is
Answer: $x=1$
Solution: $\frac{x(x+1)}{x \oplus(x+1)}=\frac{\frac{1}{x}-\frac{1}{x+1}}{\frac{1}{x}+\frac{1}{x+1}}=\frac{\frac{x+1-x}{x(x+1)}}{\frac{x+1+x}{x(x+1)}}=\frac{1}{2 x+1}$,
Therefore, the equation $\frac{x(x+1)}{x \oplus(x+1)}=\frac{1}{3}$ is $\frac{1}{2 x+1}=\frac{1}{3}$,
Solving gives
Upon verification, $\mathrm{x}=1$ is the solution to the original equation.