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Penjelasan dengan langkah-langkah:
[tex]12. \: \frac{2}{ \sqrt{6} - \sqrt{2} } = \frac{2}{ \sqrt{6} - \sqrt{2} } \times \frac{ \sqrt{6} + \sqrt{2} }{ \sqrt{6} + \sqrt{2} } \\ = \frac{2( \sqrt{6} + \sqrt{2}) }{( \sqrt{6} - \sqrt{2} )( \sqrt{6} + \sqrt{2}) } \\ = \frac{2 (\sqrt{6} + \sqrt{2}) }{6 + \sqrt{12} - \sqrt{12} - 2 } \\ = \frac{2( \sqrt{6} + \sqrt{2}) }{4} \\ = \frac{1}{2} ( \sqrt{6} + \sqrt{2} ) \: (c)[/tex]
[tex]13. \: k = {8}^{ \frac{2}{3} } \times {8}^{ \frac{1}{3} } \times {8}^{ \frac{1}{3} } \times {8}^{ \frac{2}{3} } \\ k = {8}^{ \frac{2}{3} + \frac{1}{3} + \frac{1}{3} + \frac{2}{3} } \\ k = {8}^{2} \\ k = 64 \: (bilangan \: nonprima \: dan \: lebih \: dari \: 2) \: (d) [/tex]
[tex]14. \: \frac{ {9}^{ - \frac{2}{3} }. {3}^{ - \frac{2}{3} } + {27}^{ - 1} }{ {27}^{ - \frac{ 4}{3} } + {81}^{ - \frac{3}{4} } } \\ = \frac{ {( {3}^{2}) }^{ - \frac{2}{3} }. {3}^{ - \frac{2}{3} } + { ({3}^{3}) }^{ - 1} }{ {( {3}^{3}) }^{ - \frac{4}{3} } + { ({3}^{4}) }^{ - \frac{3}{4} } } \\ = \frac{ {3}^{ - \frac{4}{3} + ( - \frac{2}{3}) } + {3}^{ - 3} }{ {3}^{ - 4} + {3}^{ - 3} } \\ = \frac{ {3}^{ - 2} + {3}^{ - 3} }{ {3}^{ - 4} + {3}^{ - 3} } \\ = \frac{ ({3}^{ - 2})(1 + {3}^{ - 1}) }{( {3}^{ - 2})( {3}^{ - 2} + {3}^{ - 1} )} \\ = \frac{1 + \frac{1}{3} }{ \frac{1}{9} + \frac{1}{3} } \\ = \frac{ \frac{4}{3} }{ \frac{4}{9} } \\ = \frac{4}{3} \times \frac{9}{4} \\ = 3 \: (c) [/tex]
[tex]15. \: panjang \: diagonal = \sqrt{ {p}^{2} + {l}^{2} } \\ = \sqrt{ {(2 + \sqrt{3} )}^{2} + { (\sqrt{3 } - 2) }^{2} } \\ = \sqrt{4 + 4 \sqrt{3} + 3 + 3 - 4 \sqrt{3} + 4} \\ = \sqrt{14} \: cm \: (c)[/tex]
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