2.Rasioanalkan
a. 4/√6-√3 =
b. 2√2/3√3-√5 =
1
[tex] \sqrt{147} + \sqrt{75} - \sqrt{48} \\ = \sqrt{49 \times 3} + \sqrt{25 \times 3} - \sqrt{16 \times 3} \\ = 7 \sqrt{3} + 5 \sqrt{3} - 4 \sqrt{3} \\ = 8 \sqrt{3} [/tex]
2a
[tex] \frac{4}{ \sqrt{6} - \sqrt{3} } = \frac{4}{ \sqrt{6} - \sqrt{3} } \times \frac{ \sqrt{6} + \sqrt{3} }{ \sqrt{6} + \sqrt{3} } \\ = \frac{4 \sqrt{6} + 4 \sqrt{3} }{6 + \sqrt{18} - \sqrt{18} - 3} \\ = \frac{4 \sqrt{6} + 4 \sqrt{3} }{3} [/tex]
2b
[tex] \frac{2 \sqrt{2} }{3 \sqrt{3} - \sqrt{5} } = \frac{2 \sqrt{2} }{3 \sqrt{3} - \sqrt{5} } \times \frac{3 \sqrt{3} + \sqrt{5} }{3 \sqrt{3} + \sqrt{5} } \\ = \frac{6 \sqrt{6} + 2 \sqrt{10} }{27 - 5} \\ = \frac{6 \sqrt{6} + 2 \sqrt{10} }{22} \\ = \frac{2(3 \sqrt{6} + \sqrt{10}) }{22} \\ = \frac{3 \sqrt{6} + \sqrt{10} }{11} [/tex]
[answer.2.content]
