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kolezko [41]
2 years ago
14

Conjugate/Rational Number?

Mathematics
1 answer:
hram777 [196]2 years ago
8 0

Answer:

1)  \dfrac{2}{\sqrt{5} }  = \dfrac{2 \cdot \sqrt{5} }{5}

2)  -\dfrac{5}{\sqrt{3} } = -\dfrac{5 \cdot \sqrt{3} }{3}

3)  \dfrac{\sqrt{2} + \sqrt{5}  }{\sqrt{10} } =\dfrac{\sqrt{5}  }{5} + \dfrac{ \sqrt{2}  }{2}

4)  \dfrac{3 + \sqrt{2} }{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = \sqrt{3} + \dfrac{\sqrt{6} }{3}

5)  \dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2}  }= \dfrac{\sqrt{15} - \sqrt{6} }{3}

Step-by-step explanation:

The rationalization of the denominator of the surds are found as follows;

1) \dfrac{2}{\sqrt{5} }

\dfrac{2}{\sqrt{5} } \times \dfrac{\sqrt{5} }{\sqrt{5} } = \dfrac{2 \cdot \sqrt{5} }{5}

\dfrac{2}{\sqrt{5} }  = \dfrac{2 \cdot \sqrt{5} }{5}

2) -\dfrac{5}{\sqrt{3} }

-\dfrac{5}{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = -\dfrac{5 \cdot \sqrt{3} }{3}

-\dfrac{5}{\sqrt{3} } = -\dfrac{5 \cdot \sqrt{3} }{3}

3) \dfrac{\sqrt{2} + \sqrt{5}  }{\sqrt{10} }

\dfrac{\sqrt{2} + \sqrt{5}  }{\sqrt{10} } \times \dfrac{ \sqrt{10}  }{\sqrt{10} } = \dfrac{\sqrt{20} + \sqrt{50}  }{10 } = \dfrac{2\cdot \sqrt{5} + 5 \cdot \sqrt{2}  }{10} = \dfrac{\sqrt{5}  }{5} + \dfrac{ \sqrt{2}  }{2}

\dfrac{\sqrt{2} + \sqrt{5}  }{\sqrt{10} } =\dfrac{\sqrt{5}  }{5} + \dfrac{ \sqrt{2}  }{2}

4) \dfrac{3 + \sqrt{2} }{\sqrt{3} }

\dfrac{3 + \sqrt{2} }{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = \dfrac{3 \cdot \sqrt{3}+\sqrt{6}  }{3 } = \sqrt{3} + \dfrac{\sqrt{6} }{3}

\dfrac{3 + \sqrt{2} }{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = \sqrt{3} + \dfrac{\sqrt{6} }{3}

5) \dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2}  }

\dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2}  } = \dfrac{\sqrt{5} - \sqrt{2} }{\sqrt{5} - \sqrt{2} }  = \dfrac{\sqrt{15} -\sqrt{6} }{5 - 2} = \dfrac{\sqrt{15} - \sqrt{6} }{3}

\dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2}  }= \dfrac{\sqrt{15} - \sqrt{6} }{3}

6) \dfrac{\sqrt{7} }{\sqrt{3} - \sqrt{5}  }

\dfrac{\sqrt{7} }{\sqrt{3} - \sqrt{5}  } \times \dfrac{\sqrt{3} + \sqrt{5}}{\sqrt{3} + \sqrt{5}}  = \dfrac{\sqrt{21} + \sqrt{35}}{{3} + {5}} = \dfrac{\sqrt{21} + \sqrt{35}}{8}

\dfrac{\sqrt{7} }{\sqrt{3} - \sqrt{5}  } \times \dfrac{\sqrt{3} + \sqrt{5}}{\sqrt{3} + \sqrt{5}}  =\dfrac{\sqrt{21} + \sqrt{35}}{8}

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