Notice that $\sqrt{51}\approx 7+\frac{\sqrt{2}}{10}$ I was doing some other stuff and noticed that:
$$\sqrt{51} - \left(7 + \dfrac{\sqrt{2}}{10}\right) = 0.0000070723\,\, (*)$$
and this immediately made me think of their respective continued fractions, which are,
$$\sqrt{51} = 7 + \dfrac{1}{7+\dfrac{1}{14+\dfrac{1}{7+\dfrac{1}{14+\dots}}}} = [7;7,14]$$
and
$$\sqrt{2} = 1+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\dots}}} = [1;2].$$
But I do not an immediate way to derive the above approximation based on these continued fractions and so I was wondering if there is an easy trick here.
For what it's worth, when  you repeatedly square $(*)$ to get rid of the square roots, it amounts to:
$$4999^2-51\cdot 700^2 = 1.$$
But they are not the smallest solutions to the above Pell's equation since
$$50^2 - 51*7^2 = 1.$$
So I guess my question is can we derive $(*)$ from some known identity/approximations? Also,  I would be curious to see if someone know of some other near-identites involving small numbers like $(*).$

To approach with continued fractions:
$$\begin{align}a_n=\sqrt{n^2+2}&=[n,n,2n,n,2n,\dots]\\\sqrt{n^2+1}&=[n,2n,2n,\dots]\\
b_n=n+\frac{1}{\sqrt{n^2+1}}&=[n,n,2n,2n,2n,\dots]
\end{align}$$
So the third continued fraction for both $a_n$ and $b_n$ are $$c_3=\frac{2n^3+3n}{2n^2+1}.$$ The fourth fraction for $a_n$ is: $$c_4=\frac{2n^4+4n^2+1}{2n^3+2n}.$$ We know $c_3<b_n<a_n<c_4,$  so $$0<a_n-b_n<c_4-c_3=\frac{1}{(2n^3+2n)(2n^2+1)}<\frac{1}{4n^5}$$
Your case is $n=7,$ because $\frac{\sqrt2}{10}=\frac1{\sqrt{7^2+1}}.$
If you have solutions of $n^2-2m^2=-1,$ you get the approximation: $$\sqrt{n^2+2}\approx n+\frac{\sqrt{2}}{2m}$$ There are infinitely many integer solutions $(n,m)$ computed as $n+m\sqrt2=(1+\sqrt2)(3+\sqrt2)^k.$
For example, $(n.m)=(7,5)$ or $(n,m)=(41,29).$ The latter gives:
$$\sqrt{1683}-41-\frac{\sqrt2}{58}=0.000000001077321\approx\frac12\cdot\frac{1}{4\cdot 41^5}$$
That $1/2$ is probably because the fourth coefficients of $a_n$ and $b_n$ are $n$ and $2n,$ respectively.

A little more work shows that $$a_n-b_n\sim\frac1{8n^5}$$
Specifically, I think you get: $$0<\frac{1}{2n(n^2+1)(4n^2+3)}-\left(a_n-b_n\right)<\frac{1}{32n^7}$$
