Tolong dong gaes, mau make sure aja jawaban saya benar atau tidak
Answer (1)
Penyelesaiana. [tex](\displaystyle \frac{3\sqrt{2}}{\sqrt{3}})[/tex][tex]=\frac{3\sqrt{2}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}[/tex][tex]= \frac{3\sqrt{6}}{3}[/tex][tex]= \sqrt{6}[/tex]b.[tex](\displaystyle \frac{3+\sqrt{2}}{2\sqrt{3}})[/tex][tex]=\frac{3+\sqrt{2}}{2\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}[/tex][tex]= \frac{3\sqrt{3} + \sqrt{6}}{2 \cdot 3}[/tex][tex]= \frac{3\sqrt{3} + \sqrt{6}}{6}[/tex]c. [tex](\displaystyle \frac{3-\sqrt{2}}{3+\sqrt{2}})[/tex][tex]=\frac{3-\sqrt{2}}{3+\sqrt{2}} \cdot \frac{3-\sqrt{2}}{3-\sqrt{2}}[/tex][tex]= \frac{(3-\sqrt{2})^2}{(3+\sqrt{2})(3-\sqrt{2})}[/tex][tex](3-\sqrt{2})^2 = 9 - 6\sqrt{2} + 2[/tex][tex]= 11 - 6\sqrt{2}[/tex][tex]= \frac{11 - 6\sqrt{2}}{7}[/tex]d. [tex](\displaystyle \frac{3\sqrt{3}}{\sqrt{3}-\sqrt{2}})[/tex][tex]=\frac{3\sqrt{3}}{\sqrt{3}-\sqrt{2}} \cdot \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}[/tex][tex]= \frac{3\sqrt{3}(\sqrt{3}+\sqrt{2})}{3-2}[/tex][tex](3\sqrt{3}\cdot\sqrt{3} + 3\sqrt{3}\cdot\sqrt{2} = 9 + 3\sqrt{6})[/tex][tex]= 9 + 3\sqrt{6}[/tex]
