7. In the equation below, all denominators are four-digit numbers. Please fill in each box with a digit to make the equation true.



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7. $\frac{1}{5964}+\frac{1}{1998}=\frac{1}{1491}$ or $\frac{1}{4970}+\frac{1}{1998}=\frac{1}{1420}$
7.【Solution】 $\frac{1}{12}+\frac{1}{4}=\frac{1}{3}$
Since $1988=2 \times 2 \times 7 \times 71=4 \times 497$, multiplying both sides of the above equation by $\frac{1}{497}$ gives $\frac{1}{5964}+\frac{1}{1988}=\frac{1}{1491}$
This provides one set of solutions that meet the conditions.
For example, $\frac{1}{35}+\frac{1}{14}=\frac{1}{10}$, multiplying both sides by $\frac{1}{142}$ gives $\frac{1}{4970}+\frac{1}{1988}=\frac{1}{1420}$
This provides another set of solutions.