Matematika

Pertanyaan

bentu sederhana pecahan
bentu sederhana pecahan

1 Jawaban


  • [tex] = \frac{ \sqrt{3} }{ \sqrt{6} - \sqrt{2} } \times \frac{ \sqrt{6} + \sqrt{2} }{ \sqrt{6} + \sqrt{2} } \\ \\ = \frac{ \sqrt{18} + \sqrt{6} }{6 - 2} \\ \\ = \frac{ \sqrt{9 \times 2} + \sqrt{6} }{4} \\ \\ = \frac{3 \sqrt{2} + \sqrt{6} }{4} \\ \\ = \frac{3}{4} \sqrt{2} + \frac{1}{4} \sqrt{6} [/tex]