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\frac{\sqrt{3}}{2\sqrt{3}+\sqrt{6}}
Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.
\frac{\sqrt{3}\left(2\sqrt{3}-\sqrt{6}\right)}{\left(2\sqrt{3}+\sqrt{6}\right)\left(2\sqrt{3}-\sqrt{6}\right)}
Rationalize the denominator of \frac{\sqrt{3}}{2\sqrt{3}+\sqrt{6}} by multiplying numerator and denominator by 2\sqrt{3}-\sqrt{6}.
\frac{\sqrt{3}\left(2\sqrt{3}-\sqrt{6}\right)}{\left(2\sqrt{3}\right)^{2}-\left(\sqrt{6}\right)^{2}}
Consider \left(2\sqrt{3}+\sqrt{6}\right)\left(2\sqrt{3}-\sqrt{6}\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{3}\left(2\sqrt{3}-\sqrt{6}\right)}{2^{2}\left(\sqrt{3}\right)^{2}-\left(\sqrt{6}\right)^{2}}
Expand \left(2\sqrt{3}\right)^{2}.
\frac{\sqrt{3}\left(2\sqrt{3}-\sqrt{6}\right)}{4\left(\sqrt{3}\right)^{2}-\left(\sqrt{6}\right)^{2}}
Calculate 2 to the power of 2 and get 4.
\frac{\sqrt{3}\left(2\sqrt{3}-\sqrt{6}\right)}{4\times 3-\left(\sqrt{6}\right)^{2}}
The square of \sqrt{3} is 3.
\frac{\sqrt{3}\left(2\sqrt{3}-\sqrt{6}\right)}{12-\left(\sqrt{6}\right)^{2}}
Multiply 4 and 3 to get 12.
\frac{\sqrt{3}\left(2\sqrt{3}-\sqrt{6}\right)}{12-6}
The square of \sqrt{6} is 6.
\frac{\sqrt{3}\left(2\sqrt{3}-\sqrt{6}\right)}{6}
Subtract 6 from 12 to get 6.
\frac{2\left(\sqrt{3}\right)^{2}-\sqrt{3}\sqrt{6}}{6}
Use the distributive property to multiply \sqrt{3} by 2\sqrt{3}-\sqrt{6}.
\frac{2\times 3-\sqrt{3}\sqrt{6}}{6}
The square of \sqrt{3} is 3.
\frac{6-\sqrt{3}\sqrt{6}}{6}
Multiply 2 and 3 to get 6.
\frac{6-\sqrt{3}\sqrt{3}\sqrt{2}}{6}
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{6-3\sqrt{2}}{6}
Multiply \sqrt{3} and \sqrt{3} to get 3.