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\left(\sqrt{2}\right)^{2}+2\sqrt{2}\sqrt{3}+\left(\sqrt{3}\right)^{2}-\sqrt{24}
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{2}+\sqrt{3}\right)^{2}.
2+2\sqrt{2}\sqrt{3}+\left(\sqrt{3}\right)^{2}-\sqrt{24}
The square of \sqrt{2} is 2.
2+2\sqrt{6}+\left(\sqrt{3}\right)^{2}-\sqrt{24}
To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.
2+2\sqrt{6}+3-\sqrt{24}
The square of \sqrt{3} is 3.
5+2\sqrt{6}-\sqrt{24}
Add 2 and 3 to get 5.
5+2\sqrt{6}-2\sqrt{6}
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}.
5
Combine 2\sqrt{6} and -2\sqrt{6} to get 0.