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\frac{\sqrt{2}}{\sqrt{3}}-\left(\frac{1}{6}\sqrt{24}-\frac{3}{2}\sqrt{12}\right)-1
Rewrite the square root of the division \sqrt{\frac{2}{3}} as the division of square roots \frac{\sqrt{2}}{\sqrt{3}}.
\frac{\sqrt{2}\sqrt{3}}{\left(\sqrt{3}\right)^{2}}-\left(\frac{1}{6}\sqrt{24}-\frac{3}{2}\sqrt{12}\right)-1
Rationalize the denominator of \frac{\sqrt{2}}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.
\frac{\sqrt{2}\sqrt{3}}{3}-\left(\frac{1}{6}\sqrt{24}-\frac{3}{2}\sqrt{12}\right)-1
The square of \sqrt{3} is 3.
\frac{\sqrt{6}}{3}-\left(\frac{1}{6}\sqrt{24}-\frac{3}{2}\sqrt{12}\right)-1
To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.
\frac{\sqrt{6}}{3}-\left(\frac{1}{6}\times 2\sqrt{6}-\frac{3}{2}\sqrt{12}\right)-1
Factor 24=2^{2}\times 6. Rewrite the square root of the product \sqrt{2^{2}\times 6} as the product of square roots \sqrt{2^{2}}\sqrt{6}. Take the square root of 2^{2}.
\frac{\sqrt{6}}{3}-\left(\frac{2}{6}\sqrt{6}-\frac{3}{2}\sqrt{12}\right)-1
Multiply \frac{1}{6} and 2 to get \frac{2}{6}.
\frac{\sqrt{6}}{3}-\left(\frac{1}{3}\sqrt{6}-\frac{3}{2}\sqrt{12}\right)-1
Reduce the fraction \frac{2}{6} to lowest terms by extracting and canceling out 2.
\frac{\sqrt{6}}{3}-\left(\frac{1}{3}\sqrt{6}-\frac{3}{2}\times 2\sqrt{3}\right)-1
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{6}}{3}-\left(\frac{1}{3}\sqrt{6}-3\sqrt{3}\right)-1
Cancel out 2 and 2.
\frac{\sqrt{6}}{3}-\frac{1}{3}\sqrt{6}-\left(-3\sqrt{3}\right)-1
To find the opposite of \frac{1}{3}\sqrt{6}-3\sqrt{3}, find the opposite of each term.
-\left(-3\sqrt{3}\right)-1
Combine \frac{\sqrt{6}}{3} and -\frac{1}{3}\sqrt{6} to get 0.
3\sqrt{3}-1
The opposite of -3\sqrt{3} is 3\sqrt{3}.