3 years ago

What is the length of r?

Mathematics

1 answer:

muminat3 years ago

5 0

I don't understand your question. can you make it clear

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

jasenka [17]

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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A bag of marbles contains 4 red

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