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The question denotes two equations:

(1) $\frac{a + b}{2} - \frac{3}{2} = \sqrt{ab}$ (2) $\sqrt{ab} - \frac{6}{5} = \frac{2}{\frac{1}{a} + \frac{1}{b}}$

Simplifying equation 2 we get

$\sqrt{ab} - \frac{6}{5} = \frac{2ab}{a+b}$

Then, subtracting equation 1 from equation 2, we get $\begin{array}{lcl} \sqrt{ab} - \frac{6}{5} = \frac{2ab}{a+b} \\ -\sqrt{ab} -\frac{3}{2} = -\frac{a+b}{2} \end{array}$ $-\frac{27}{10} = \frac{2ab}{a+b} - \frac{a+b}{2}$ Simplifying and combining terms $-\frac{27}{10} = \frac{4ab}{2(a+b)} - \frac{(a+b)^{2}}{2(a+b)}$ $-\frac{27}{10} = \frac{4ab - a^{2} -2ab - b^{2}}{2(a+b)}$ $-\frac{27}{10} = \frac{- a^{2} +2ab - b^{2}}{2(a+b)}$

(3) $a^{2} + b^{2} = \frac{27}{5}a + \frac{27}{5}b +2ab$

Let's call this previous result equation 3. Now we are going to head back and manipulate equation 1 to find a suitable result. Squaring equation one and rearranging results in

(4) $a^{2} + b^{2} = 2ab +6a +6b -9$ Subtracting equation 3 from equation 4 result in $\begin{array}{lcl} a^{2} + b^{2} = 6a +6b + 2ab -9 \\ -a^{2} + -b^{2} = -\frac{27}{5}a + -\frac{27}{5}b -2ab\end{array}$ $0 = \frac{3}{5}a + \frac{3}{5}b - 9$ $a + b = 15$

Plugging this result back into equation 1 we get $\frac{15}{2} - \frac{3}{2} = \sqrt{ab}$ $6 = \sqrt{ab}$ $ab = 36$ Squaring $a + b = 15$ results in $a^2 + b^2 = 225 -2ab$ Substituting in $ab = 36$ we get $a^2 + b^2 = 225 -72 = \boxed{153}$