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Thread: [SOLVED] Fractions with surds in the denominator

  1. #1
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    [SOLVED] Fractions with surds in the denominator

    Simply by rationalising the denominator

    $\displaystyle \frac {\sqrt 3}
    {\sqrt 6}$

    thanks!
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  2. #2
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    Quote Originally Posted by waven View Post
    Simply by rationalising the denominator

    $\displaystyle \frac {\sqrt 3}
    {\sqrt 6}$

    thanks!

    multiply it by $\displaystyle \frac{\sqrt{6}}{\sqrt{6}}$
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  3. #3
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    Quote Originally Posted by mathaddict View Post
    multiply it by $\displaystyle \frac{\sqrt{6}}{\sqrt{6}}$
    $\displaystyle \frac {\sqrt 18} 6$ what do i do next?

    btw the answer is $\displaystyle \frac {\sqrt 2} {2}$
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  4. #4
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    Quote Originally Posted by waven View Post
    $\displaystyle \frac {\sqrt 18} 6$ what do i do next?

    btw the answer is $\displaystyle \frac {\sqrt 2} {2}$
    Notice that $\displaystyle \sqrt{18} = \sqrt{9 \times 2} = \sqrt{9} \times \sqrt{2} = 3\sqrt{2}$.

    So $\displaystyle \frac{\sqrt{18}}{6} = \frac{3\sqrt{2}}{6} = \frac{\sqrt{2}}{2}$.
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