Hasil dari 2+2√6 / √8-√3

Operasi Bentuk Akar

[tex] \frac{2 + 2 \sqrt{6} }{ \sqrt{8} - \sqrt{3} } = \frac{2 + 2 \sqrt{6} }{ \sqrt{8} - \sqrt{3} } \times \frac{ \sqrt{8} + \sqrt{3} }{ \sqrt{8} + \sqrt{3} } \\ \\ = \frac{(2 + 2 \sqrt{6}) ( \sqrt{8} + \sqrt{3}) }{8 - 3} \\ \\ = \frac{2 \sqrt{8} + 2 \sqrt{3} + 2 \sqrt{48} + 2 \sqrt{18} }{5} \\ \\ = \frac{2 \sqrt{4 \times 2} + 2 \sqrt{3} + 2 \sqrt{16 \times 3} + 2 \sqrt{9 \times 2} }{5} \\ \\ = \frac{4 \sqrt{2} + 2 \sqrt{3} + 8 \sqrt{3} + 6 \sqrt{2} }{5} \\ \\ = \frac{10 \sqrt{2} + 10 \sqrt{3} }{5} \\ \\ = 2 \sqrt{2 } + 2 \sqrt{3} [/tex]


*semoga membantu. sukses selalu ^_^