(3)/(sqrt(3)+sqrt(2))-(2)/(sqrt(3)-sqrt(2))

To solve the equation $\frac {3}{\sqrt {3+\sqrt {2}}}=\frac {2}{\sqrt {3}-\sqrt {2}}$, we need to simplify both sides of the equation and see if they are equal.<br /><br />Let's start with the left side of the equation. We can rationalize the denominator by multiplying both the numerator and denominator by $\sqrt {3+\sqrt {2}}$:<br /><br />$\frac {3}{\sqrt {3+\sqrt {2}}} \cdot \frac {\sqrt {3+\sqrt {2}}}{\sqrt {3+\sqrt {2}}} = \frac {3\sqrt {3+\sqrt {2}}}{3+\sqrt {2}}$<br /><br />Now let's simplify the right side of the equation. We can also rationalize the denominator by multiplying both the numerator and denominator by $\sqrt {3+\sqrt {2}}$:<br /><br />$\frac {2}{\sqrt {3}-\sqrt {2}} \cdot \frac {\sqrt {3}+\sqrt {2}}{\sqrt {3}+\sqrt {2}} = \frac {2(\sqrt {3}+\sqrt {2})}{3-2\sqrt {2}}$<br /><br />Now we can simplify both sides of the equation further:<br /><br />Left side: $\frac {3\sqrt {3+\sqrt {2}}}{3+\sqrt {2}}$<br /><br />Right side: $\frac {2(\sqrt {3}+\sqrt {2})}{3-2\sqrt {2}}$<br /><br />Since both sides of the equation are equal, the equation is true. Therefore, the correct answer is True.