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rationalize the denominator of 7/(3 + √3)
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Multiply top and bottom by 3-√3 to get \[\Large \frac{7(3-\sqrt{3})}{(3+\sqrt{3})(3-\sqrt{3})}\] \[\Large \frac{21-7\sqrt{3}}{3^2-(\sqrt{3})^2}\] \[\Large \frac{21-7\sqrt{3}}{9-3}\] \[\Large \frac{21-7\sqrt{3}}{6}\] So \[\Large \frac{7}{3+\sqrt{3}} = \frac{21-7\sqrt{3}}{6}\]
21-7 square root 3/ 6..
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