Matematika Sekolah Menengah Atas jika p= 5+√3 dan q=5-√3 tentukan=1. p-q/q

davionwunschulv – April 11, 2023

jika p= 5+√3 dan q=5-√3 tentukan=

1. p-q/q

[tex]\frac{(5 + \sqrt{3}) - (5 - \sqrt{3})}{5 - \sqrt{3}} = \frac{5 + \sqrt{3} - 5 + \sqrt{3}}{5 - \sqrt{3}} = \frac{5 - 5 + \sqrt{3} + \sqrt{3}}{5 - \sqrt{3}} = \frac{2 \sqrt{3}}{5 - \sqrt{3}} [/tex]

Kemudian rasionalkan penyebutnya,

[tex]\frac{2 \sqrt{3}}{5 - \sqrt{3}} = \frac{2 \sqrt{3}}{5 - \sqrt{3}} \times \frac{5 + \sqrt{3}}{5 + \sqrt{3}} = \frac{(2 \sqrt{3})(5 + \sqrt{3})}{(5 - \sqrt{3})(5 + \sqrt{3})} [/tex]

[tex]= \frac{10 \sqrt{3} + 2 \sqrt{9}}{25 + 5 \sqrt{3} - 5 \sqrt{3} - \sqrt{9}} [/tex]

[tex]= \frac{10 \sqrt{3} + 2 \sqrt{9}}{25 - \sqrt{9}} = \frac{10 \sqrt{3} + 2 \times 3}{25 - 3} = \frac{10 \sqrt{3} + 6}{22} = \frac{5 \sqrt{3} + 3}{11} [/tex]

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