7. Given positive integers $a, b$ satisfy
$$
\sqrt{\frac{a b}{2 b^{2}-a}}=\frac{a+2 b}{4 b} \text {. }
$$

Then $|10(a-5)(b-15)|+2=$
7.2012.

Notice,
$$
\begin{array}{l}
\sqrt{\frac{a b}{2 b^{2}-a}}=\frac{a+2 b}{4 b} \\
\Leftrightarrow 16 a b^{3}=\left(2 b^{2}-a\right)\left(a^{2}+4 a b+4 b^{2}\right) \\
\Leftrightarrow a\left(a^{2}+4 a b+4 b^{2}\right)=2 b^{2}\left(a^{2}-4 a b+4 b^{2}\right) \\
\Leftrightarrow a(a+2 b)^{2}=2 b^{2}(a-2 b)^{2} .
\end{array}
$$

Let $d=(a, b), a=a_{0} d, b=b_{0} d,\left(a_{0}, b_{0}\right)=1$.
Then equation (1) $\Leftrightarrow a_{0}\left(a_{0}+2 b_{0}\right)^{2}=2 d b_{0}^{2}\left(a_{0}-2 b_{0}\right)^{2}$.
Since $b_{0} \mid a_{0}\left(a_{0}+2 b_{0}\right)^{2}$, we get
$$
b_{0} \mid a_{0}^{3} \Rightarrow b_{0}=1 \text {. }
$$

Then $a_{0}\left(a_{0}+2\right)^{2}=2 d\left(a_{0}-2\right)^{2}$
$$
\Rightarrow 2 d=a_{0}+\frac{8 a_{0}^{2}}{\left(a_{0}-2\right)^{2}}=a_{0}+2\left(\frac{2 a_{0}}{a_{0}-2}\right)^{2} \text {. }
$$

Thus $\left(a_{0}-2\right)\left|2 a_{0} \Rightarrow\left(a_{0}-2\right)\right| 4$
$$
\Rightarrow a_{0}-2= \pm 1,2,4 \text {. }
$$

From equation (2), we know $2 \mid a_{0}$. Therefore, $a_{0}=4,6$.
Substituting into equation (2) gives $d=18,12$.
Accordingly, $(a, b)=(72,18),(72,12)$.
Thus, $|10(a-5)(b-15)|+2=2012$.