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We want to square the equation in order to eliminate the radicals. To do so, we first move the $\sqrt{x+\frac{4}{x}}$ term to the right-hand side, giving $$\sqrt{x}+\sqrt{\frac{4}{x}}=6-\sqrt{x+\frac{4}{x}}.$$Now we see that squaring will produce lots of common terms on the left-hand and right-hand sides, which cancel: $$\begin{array}{rl}& \\ {\left(\sqrt{x}+\sqrt{\frac{4}{x}}\right)}^2& ={\left(6-\sqrt{x+\frac{4}{x}}\right)}^2\\ x+4+\frac{4}{x}& =36-12\sqrt{x+\frac{4}{x}}+\left(x+\frac{4}{x}\right)\end{array}$$which simplifies to $3\sqrt{x+\frac{4}{x}}=8.$ Squaring both sides, multiplying, and rearranging gives the quadratic $$9x^2-64x+36=0.$$By Vieta's formulas, the sum of the roots of this quadratic is $\boxed{\frac{64}{9}}.$

To be complete, we must check that both of these roots satisfy the original equation. There are two steps in our above solution which could potentially not be reversible: squaring the equation $$\sqrt{x}+\sqrt{\frac{4}{x}}=6-\sqrt{x+\frac{4}{x}},$$and squaring the equation $$3\sqrt{x+\frac{4}{x}}=8.$$To check that these steps are reversible, we need to make sure that both sides of the equations in both steps are nonnegative whenever $x$ is a root of $9x^2-64x+36=0.$ This quadratic is equivalent to $x+\frac{4}{x}=\frac{64}{9},$ so $6-\sqrt{x+\frac{4}{x}}=6-\sqrt{\frac{64}{9}}=\frac{10}{3},$ which is positive, and $3\sqrt{x+\frac{4}{x}}=3\sqrt{\frac{64}{9}}=8,$ which is also positive. Therefore, all our steps were reversible, so both roots of the quadratic satisfy the original equation as well.