7. Given $a=2 \sin 45^{\circ}+1, b=2 \cos 45^{\circ}-1$. Then the value of the algebraic expression $\left(\frac{a^{2}+b^{2}}{2 a b}-1\right) \div\left(\frac{a^{2}-b^{2}}{a^{2} b+a b^{2}}\right)$ is
\[\begin{array}{l}\left(\frac{a^{2}+b^{2}}{2 a b}-1\right) \div\left(\frac{a^{2}-b^{2}}{a^{2} b+a b^{2}}\right) \\ =\frac{(a-b)^{2}}{2 a b} \cdot \frac{a b(a+b)}{(a+b)(a-b)} \\ =\frac{a-b}{2}=1\end{array}\]