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Matematika tolong merasionalkan bentuk akar dibawah ini kak buat bsokk​

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tolong merasionalkan bentuk akar dibawah ini kak buat bsokk​

Rasionalkan bentuk akar di bawah ini:

  • [tex]\frac{10}{3+\sqrt{3} }[/tex]
  • [tex]\frac{\sqrt{10} }{4-\sqrt{2} }[/tex]
  • [tex]\frac{2\sqrt{5} }{5+\sqrt{3} }[/tex]
  • [tex]\frac{4\sqrt{6} }{\sqrt{3} -\sqrt{2}}}[/tex]
  • [tex]\frac{4+\sqrt{2} }{2\sqrt{3}+3\sqrt{3} }[/tex]

Jawaban soal

  • [tex]\frac{10}{3+\sqrt{3} } =5 - \frac{5}{3} \sqrt{3}[/tex]
  • [tex]\frac{\sqrt{10} }{4-\sqrt{2} } = \frac{2+4\sqrt{2} }{7}[/tex]
  • [tex]\frac{2\sqrt{5} }{5+\sqrt{3} } = \frac{5\sqrt{5} -\sqrt{15} }{16 }[/tex]
  • [tex]\frac{4\sqrt{6} }{\sqrt{3} -\sqrt{2}}} =16 \sqrt{2}-12 \sqrt{4}[/tex]
  • [tex]\frac{4+\sqrt{2} }{2\sqrt{3}+3\sqrt{3} } = \frac{4\sqrt{3}-\sqrt{6} }{9}[/tex]

Penjelasan dan langkah-langkah

Diketahui:

Bentuk akar dari:

  • [tex]\frac{10}{3+\sqrt{3} }[/tex]
  • [tex]\frac{\sqrt{10} }{4-\sqrt{2} }[/tex]
  • [tex]\frac{2\sqrt{5} }{5+\sqrt{3} }[/tex]
  • [tex]\frac{4\sqrt{6} }{\sqrt{3} -\sqrt{2}}}[/tex]
  • [tex]\frac{4+\sqrt{2} }{2\sqrt{3}+3\sqrt{3} }[/tex]

Ditanyakan:

  • Merasionalkan bentuk akar = ?

Jawab:

  • [tex]\frac{10}{3+\sqrt{3} }[/tex]

[tex]\frac{10}{3+\sqrt{3} }\\= \frac{10}{3+\sqrt{3}} . \frac{3-\sqrt{3}}{3-\sqrt{3}} \\\\= \frac{10 (3-\sqrt{3})}{3^{2} -(\sqrt{3})^2}\\\\=\frac{10 (3-\sqrt{3})}{9 -3}\\\\=\frac{30-10\sqrt{3}}{6}\\=5 - \frac{5}{3} \sqrt{3}[/tex]

  • [tex]\frac{\sqrt{10} }{4-\sqrt{2} }[/tex]

[tex]\frac{\sqrt{10} }{4-\sqrt{2} }\\=\frac{2\sqrt{2} }{4-\sqrt{2} } x \frac{4+\sqrt{2} }{4+\sqrt{2} }\\\\= \frac{2\sqrt{2} (4+\sqrt{2})}{4^2-(\sqrt{2} )^2} \\\\= \frac{8\sqrt{2} +2.2}{16-2} \\\\= \frac{8\sqrt{2} +4}{14} \\=\frac{2+4\sqrt{2} }{7}[/tex]

  • [tex]\frac{2\sqrt{5} }{5+\sqrt{3} }[/tex]

[tex]\frac{2\sqrt{5} }{5+\sqrt{3} }\\=\frac{2\sqrt{5} }{5+\sqrt{3} } . \frac{5-\sqrt{3} }{5-\sqrt{3} }\\\\=\frac{2\sqrt{5} (5-\sqrt{3}) }{5^2+(\sqrt{3})^2 } \\\\=\frac{10\sqrt{5} -2\sqrt{15} }{25+3 }\\\\=\frac{10\sqrt{5} -2\sqrt{15} }{28 }\\= \frac{5\sqrt{5} -\sqrt{15} }{16 }[/tex]

  • [tex]\frac{4\sqrt{6} }{\sqrt{3} -\sqrt{2}}}[/tex]

[tex]\frac{4\sqrt{6} }{\sqrt{3} -\sqrt{2}}}\\=\frac{4\sqrt{6} }{\sqrt{3} -\sqrt{2}}} . \frac{\sqrt{3} +\sqrt{2}}{\sqrt{3}+-\sqrt{2}}}\\\\=\frac{4\sqrt{6}(\sqrt{3} -\sqrt{2}) }{(\sqrt{3})^2 -(\sqrt{2})^2}}\\\\= \frac{4\sqrt{6.3} -4\sqrt{6.2} }{3 -1} \\\\=\frac{4\sqrt{18} -4\sqrt{12} }{1} \\= 4 . 4 \sqrt{2} - 4.3\sqrt{4} \\=16 \sqrt{2}-12 \sqrt{4}[/tex]

  • [tex]\frac{4+\sqrt{2} }{2\sqrt{3}+3\sqrt{3} }[/tex]

[tex]\frac{4+\sqrt{2} }{2\sqrt{3}+3\sqrt{3}}\\=\frac{4+\sqrt{2} }{2\sqrt{3}+3\sqrt{3}} . \frac{2\sqrt{3}-3\sqrt{3} }{2\sqrt{3}-3\sqrt{3}}\\\\= \frac{4+\sqrt{2}(2\sqrt{3}-3\sqrt{3}) }{(2\sqrt{3})^2-(3\sqrt{3})^2} \\\\= \frac{8\sqrt{3} -12\sqrt{3} +2\sqrt{6} -3\sqrt{6} }{4.3 - 9.3}\\ \\= \frac{-4\sqrt{3}-\sqrt{6} }{12 -21} \\\\=\frac{-4\sqrt{3}-\sqrt{6} }{- 9} \\\\= \frac{4\sqrt{3}-\sqrt{6} }{9}[/tex]

Pelajari lebih lanjut

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