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guapka [62]
2 years ago
13

Rationalise the denominator of (12)/(\sqrt(10)+\sqrt(7)+\sqrt(3))

Mathematics
1 answer:
jasenka [17]2 years ago
5 0

Answer:

\frac{12}{\sqrt{10} +\sqrt{7} +\sqrt{3} }=\frac{6\sqrt{147} +6\sqrt{63}-6\sqrt{210}  }{21 }

Step-by-step explanation:

\frac{12}{\sqrt{10} +\sqrt{7} +\sqrt{3} }

=\frac{12\left( \sqrt{10} -\left( \sqrt{7} +\sqrt{3} \right)  \right)  }{(\sqrt{10}+(\sqrt{7} +\sqrt{3 } ))( \sqrt{10} -( \sqrt{7} +\sqrt{3}))}

=\frac{12\left( \sqrt{10} -\sqrt{7} -\sqrt{3} \right)  }{\sqrt{10^2} -(\sqrt{7} +\sqrt{3})^2}

\frac{12\left( \sqrt{10} -\sqrt{7} -\sqrt{3} \right)  }{10-(10+2\sqrt{21} )} }

=\frac{12\left( \sqrt{10} -\sqrt{7} -\sqrt{3} \right)  }{-2\sqrt{21} }

=\frac{-6\left( \sqrt{10} -\sqrt{7} -\sqrt{3} \right)  }{\sqrt{21} }

=\frac{-6\left( \sqrt{10} -\sqrt{7} -\sqrt{3} \right)  }{\sqrt{21} } \times\frac{\sqrt{21} }{\sqrt{21} }

=\frac{-6\sqrt{21} \left( \sqrt{10} -\sqrt{7} -\sqrt{3} \right)  }{21 }

=\frac{-6\sqrt{210} +6\sqrt{147} +6\sqrt{63}  }{21 }

=\frac{6\sqrt{147} +6\sqrt{63}-6\sqrt{210}  }{21 }

<u><em>Remark</em></u>:

You can simplify moreover if you want to.

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