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\frac{\sqrt{6}\left(2\sqrt{3}-\sqrt{2}\right)}{\left(2\sqrt{3}+\sqrt{2}\right)\left(2\sqrt{3}-\sqrt{2}\right)}
Rationalize the denominator of \frac{\sqrt{6}}{2\sqrt{3}+\sqrt{2}} by multiplying numerator and denominator by 2\sqrt{3}-\sqrt{2}.
\frac{\sqrt{6}\left(2\sqrt{3}-\sqrt{2}\right)}{\left(2\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}
Consider \left(2\sqrt{3}+\sqrt{2}\right)\left(2\sqrt{3}-\sqrt{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.
\frac{\sqrt{6}\left(2\sqrt{3}-\sqrt{2}\right)}{2^{2}\left(\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}
Expand \left(2\sqrt{3}\right)^{2}.
\frac{\sqrt{6}\left(2\sqrt{3}-\sqrt{2}\right)}{4\left(\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}
Calculate 2 to the power of 2 and get 4.
\frac{\sqrt{6}\left(2\sqrt{3}-\sqrt{2}\right)}{4\times 3-\left(\sqrt{2}\right)^{2}}
The square of \sqrt{3} is 3.
\frac{\sqrt{6}\left(2\sqrt{3}-\sqrt{2}\right)}{12-\left(\sqrt{2}\right)^{2}}
Multiply 4 and 3 to get 12.
\frac{\sqrt{6}\left(2\sqrt{3}-\sqrt{2}\right)}{12-2}
The square of \sqrt{2} is 2.
\frac{\sqrt{6}\left(2\sqrt{3}-\sqrt{2}\right)}{10}
Subtract 2 from 12 to get 10.
\frac{2\sqrt{6}\sqrt{3}-\sqrt{6}\sqrt{2}}{10}
Use the distributive property to multiply \sqrt{6} by 2\sqrt{3}-\sqrt{2}.
\frac{2\sqrt{3}\sqrt{2}\sqrt{3}-\sqrt{6}\sqrt{2}}{10}
Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.
\frac{2\times 3\sqrt{2}-\sqrt{6}\sqrt{2}}{10}
Multiply \sqrt{3} and \sqrt{3} to get 3.
\frac{6\sqrt{2}-\sqrt{6}\sqrt{2}}{10}
Multiply 2 and 3 to get 6.
\frac{6\sqrt{2}-\sqrt{2}\sqrt{3}\sqrt{2}}{10}
Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.
\frac{6\sqrt{2}-2\sqrt{3}}{10}
Multiply \sqrt{2} and \sqrt{2} to get 2.