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Lady_Fox [76]
4 years ago
6

Simplify:

" title="\frac{1}{\sqrt{5}-\sqrt{7}+\sqrt{12}}" alt="\frac{1}{\sqrt{5}-\sqrt{7}+\sqrt{12}}" align="absmiddle" class="latex-formula">
Mathematics
2 answers:
Masja [62]4 years ago
8 0

Make use of the difference of squares identity,

a^2-b^2=(a-b)(a+b)

Let a=\sqrt5-\sqrt7 and b=\sqrt{12}. Then

\dfrac1{(\sqrt5-\sqrt7)+\sqrt{12}}\cdot\dfrac{(\sqrt5-\sqrt7)-\sqrt{12}}{(\sqrt5-\sqrt7)-\sqrt{12}}=\dfrac{(\sqrt5-\sqrt7)-\sqrt{12}}{(\sqrt5-\sqrt7)^2-(\sqrt{12})^2}=\dfrac{\sqrt5-\sqrt7-\sqrt{12}}{5-2\sqrt{35}+7-12}=-\dfrac{\sqrt5-\sqrt7-\sqrt{12}}{2\sqrt{35}}

Now multiply the numerator and denominator by √(35):

-\dfrac{\sqrt5-\sqrt7-\sqrt{12}}{2\sqrt{35}}\cdot\dfrac{\sqrt{35}}{\sqrt{35}}=-\dfrac{(\sqrt5-\sqrt7-\sqrt{12})\sqrt{35}}{70}=-\dfrac{5\sqrt7-7\sqrt5-2\sqrt{105}}{70}

Nikitich [7]4 years ago
6 0

Answer:

Multiply the numerator and denominator by the conjugate.

Exact Form:

1√5−√7+2√3

Decimal Form:

0.75267290…

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