a) [tex]\frac{2}{\sqrt{12} + \sqrt{3} } = \frac{2}{2\sqrt{3}+\sqrt{3} } = \frac{2}{3\sqrt{3} } * \frac{\sqrt{3} }{\sqrt{3} } = \frac{2\sqrt{3} }{9}[/tex]
b) [tex]\frac{\sqrt{27} - \sqrt{12} }{\sqrt{32} - \sqrt{18} } = \frac{3\sqrt{3} - 2\sqrt{3} }{4\sqrt{2} -3\sqrt{2} } = \frac{\sqrt{3} }{\sqrt{2} } * \frac{\sqrt{2} }{\sqrt{2} } = \frac{\sqrt{6} }{2}[/tex]
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Verified answer
a) [tex]\frac{2}{\sqrt{12} + \sqrt{3} } = \frac{2}{2\sqrt{3}+\sqrt{3} } = \frac{2}{3\sqrt{3} } * \frac{\sqrt{3} }{\sqrt{3} } = \frac{2\sqrt{3} }{9}[/tex]
b) [tex]\frac{\sqrt{27} - \sqrt{12} }{\sqrt{32} - \sqrt{18} } = \frac{3\sqrt{3} - 2\sqrt{3} }{4\sqrt{2} -3\sqrt{2} } = \frac{\sqrt{3} }{\sqrt{2} } * \frac{\sqrt{2} }{\sqrt{2} } = \frac{\sqrt{6} }{2}[/tex]