14. (15 points) Two simplest fractions, the difference between the larger and the smaller is $\frac{5}{6}$, the greatest common divisor of the two numerators equals the difference between the two numerators, and the least common multiple of the two numerators is 1050. Find these two simplest fractions.
【Analysis】Decompose 1050 into prime factors, $1050=2 \times 3 \times 5 \times 5 \times 7$, and get the greatest common divisor of two numbers equal to the difference of the two numerators 5, which gives the numerators as $3 \times 5 \times 5=75$ and $2 \times 7 \times 5=70$; then according to the larger minus the smaller difference is $\frac{5}{6}$, and thus find the denominators of the two fractions, and the solution can be obtained.

【Solution】Solution: Let these two simplest fractions be $\frac{\mathrm{am}}{\mathrm{bk}}$ and $\frac{\mathrm{cm}}{\mathrm{dk}}$,
$$
\begin{array}{l}
(b, d)=1 \\
(a, c)=1
\end{array}
$$
$$
(a m, b k)=1 ;(c m, d k)=1 \text { (3) }
$$

Since $m=a m-c m$, we have
$$
a-c=1 \text {. (4) }
$$

Since $[a m, c m]=1050=1 \times 2 \times 3 \times 5 \times 5 \times 7$, and combining with $(4)$, we can get:
(1) $c=14, a=15, m=5$, at this time, $\frac{75}{\mathrm{bk}}-\frac{70}{\mathrm{dk}}=\frac{5}{6}$, or $\frac{15}{\mathrm{bk}}-\frac{14}{\mathrm{dk}}=\frac{1}{6}$;
(2) $c=6, a=7, m=5 \times 5$, at this time, $\frac{7 \times 5}{\mathrm{bk}}-\frac{6 \times 5}{\mathrm{dk}}=\frac{1}{6}$;
(3) $c=5, a=6, m=5 \times 7$, at this time, $\frac{6 \times 7}{b k}-\frac{5 \times 7}{d k}=\frac{1}{6}$;
(4) $c=2, a=3, m=5 \times 5 \times 7$, at this time, $\frac{3 \times 5 \times 7}{b k}-\frac{2 \times 5 \times 7}{d k}=\frac{1}{6}$ (8)
(5) $c=1, a=2, m=3 \times 5 \times 5 \times 7$, at this time, $\frac{2 \times 3 \times 5 \times 7}{\mathrm{bk}}-\frac{3 \times 5 \times 7}{\mathrm{dk}}=\frac{1}{6}$

In the above (6), $\frac{7 \times 5}{\mathrm{bk}}-\frac{6 \times 5}{\mathrm{dk}}=5 \times\left(\frac{7}{\mathrm{bk}}-\frac{6}{\mathrm{dk}}\right)=\frac{1}{6}$,
combined with condition (1), there must be $5 \mid k$, i.e., $k$ has a factor of 5, which contradicts (3). Thus, $\frac{15}{\mathrm{bk}}-\frac{14}{d k}=\frac{1}{6}$ has no solution.
Similarly, (7), (8), and (9) all have $7 \mid k$, which contradicts (3), so they all have no solution.
Only consider (5),
$$
\frac{15}{b k}-\frac{14}{d k}=\frac{1}{6}, \frac{15}{b k}-\frac{14}{d k}=\frac{15 d-14 b}{k b d}=\frac{1}{\frac{k b d}{15 d-14 b}}=\frac{1}{6}
$$

According to (1), (2), and (3), there should be
$$
(b, 15 d-14 b)=1,(d, 15 d-14 b)=1,
$$

This means:
$$
k=(15 d-14 b) \times n,(11)
$$

And (10) transforms to $\frac{1}{n b d}=\frac{1}{1 \times 2 \times 3}$, i.e., $n, b, d$ can only take $1,2,3,6$.
By (3) and (11), we know: $(n, 15)=1,(n, 14)=1$, so we get $n=1$. Similarly, $(b, 15)=1$, $(d, 14)=1$, so we get: $b=2, d=3$. Therefore,
$$
b k=2 \times(15 d-14 b)=34, d k=3 \times(15 d-14 b)=51 .
$$

These two fractions are $\frac{75}{34}$ and $\frac{70}{51}$.
Answer: These two fractions are $\frac{75}{34}$ and $\frac{70}{51}$.