Answer: G7. (2√3)/3 G8. -2+√7 G9. (6 +2√2 -3√3 -√6)/7Step-by-step explanation:G7.[tex]\dfrac{2}{\sqrt{3}}=\dfrac{2}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{2\sqrt{3}}{3}[/tex]__G8.[tex]\dfrac{3}{2 +\sqrt{7}}=\dfrac{3}{2+\sqrt{7}}\cdot\dfrac{2-\sqrt{7}}{2-\sqrt{7}}=\dfrac{6-3\sqrt{7}}{2^2-(\sqrt{7})^2}\\\\=\dfrac{6-3\sqrt{7}}{-3}=\sqrt{7}-2[/tex]__G9.[tex]\dfrac{2-\sqrt{3}}{3-\sqrt{2}}=\dfrac{2-\sqrt{3}}{3-\sqrt{2}}\cdot\dfrac{3+\sqrt{2}}{3+\sqrt{2}}=\dfrac{(2-\sqrt{3})(3+\sqrt{2})}{9-2}\\\\=\dfrac{6+2\sqrt{2}-3\sqrt{3}-\sqrt{6}}{7}[/tex]_____Comment on the problemsIn most cases, these expressions are the simplest possible (take the least amount of ink to draw, and take the fewest math operations to evaluate). What seems to be intended is that the denominator be made a rational number. This is done by multiplying the given fraction by a fraction equal to 1 that has the same denominator but with the sign of the radical reversed (unless, as in the first case, the radical is by itself). The purpose of doing this is to take advantage of the fact that (a-b)(a+b) = a²-b², so if "a" or "b" is a square root, that root will not be seen in the product. In problem G9, we see this can make the numerator quite messy--not exactly a simpler form--but all the irrational numbers are in the numerator.