Solution 3.1:7a

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Current revision (14:05, 30 September 2008) (edit) (undo)
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First, we multiply the tops and bottoms of the two terms by the conjugate of their respective denominators, so that there are no root signs left in the denominators,
First, we multiply the tops and bottoms of the two terms by the conjugate of their respective denominators, so that there are no root signs left in the denominators,
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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\frac{1}{\sqrt{6}-\sqrt{5}}
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& \frac{1}{\sqrt{6}-\sqrt{5}}=\frac{1}{\sqrt{6}-\sqrt{5}}\centerdot \frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}=\frac{\sqrt{6}+\sqrt{5}}{\left( \sqrt{6} \right)^{2}-\left( \sqrt{5} \right)^{2}}=\frac{\sqrt{6}+\sqrt{5}}{6-5}=\sqrt{6}+\sqrt{5}, \\
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&= \frac{1}{\sqrt{6}-\sqrt{5}}\cdot \frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}\\[5pt]
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& \frac{1}{\sqrt{7}-\sqrt{6}}=\frac{1}{\sqrt{7}-\sqrt{6}}\centerdot \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}=\frac{\sqrt{7}+\sqrt{6}}{\left( \sqrt{7} \right)^{2}-\left( \sqrt{6} \right)^{2}}=\frac{\sqrt{7}+\sqrt{6}}{7-6}=\sqrt{7}+\sqrt{6}, \\
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&= \frac{\sqrt{6}+\sqrt{5}}{(\sqrt{6})^{2}-(\sqrt{5})^{2}}\\[5pt]
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& \\
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&= \frac{\sqrt{6}+\sqrt{5}}{6-5}\\[5pt]
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\end{align}</math>
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&= \sqrt{6}+\sqrt{5}\,,\\[10pt]
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\frac{1}{\sqrt{7}-\sqrt{6}}
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&= \frac{1}{\sqrt{7}-\sqrt{6}}\cdot \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}\\[5pt]
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&= \frac{\sqrt{7}+\sqrt{6}}{(\sqrt{7})^{2}-(\sqrt{6})^{2}}\\[5pt]
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&= \frac{\sqrt{7}+\sqrt{6}}{7-6}\\[5pt]
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&= \sqrt{7}+\sqrt{6}\,\textrm{.}
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\end{align}</math>}}
Now, we can subtract the terms and simplify the result,
Now, we can subtract the terms and simplify the result,
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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\frac{1}{\sqrt{6}-\sqrt{5}}-\frac{1}{\sqrt{7}-\sqrt{6}}
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& \frac{1}{\sqrt{6}-\sqrt{5}}-\frac{1}{\sqrt{7}-\sqrt{6}}=\sqrt{6}+\sqrt{5}-\left( \sqrt{7}+\sqrt{6} \right) \\
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&= \sqrt{6}+\sqrt{5}-(\sqrt{7}+\sqrt{6})\\[5pt]
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& =\sqrt{6}+\sqrt{5}-\sqrt{7}-\sqrt{6}=\sqrt{5}-\sqrt{7}. \\
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&= \sqrt{6}+\sqrt{5}-\sqrt{7}-\sqrt{6}\\[5pt]
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\end{align}</math>
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&= \sqrt{5}-\sqrt{7}\,\textrm{.}
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\end{align}</math>}}

Current revision

First, we multiply the tops and bottoms of the two terms by the conjugate of their respective denominators, so that there are no root signs left in the denominators,

\displaystyle \begin{align}

\frac{1}{\sqrt{6}-\sqrt{5}} &= \frac{1}{\sqrt{6}-\sqrt{5}}\cdot \frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}\\[5pt] &= \frac{\sqrt{6}+\sqrt{5}}{(\sqrt{6})^{2}-(\sqrt{5})^{2}}\\[5pt] &= \frac{\sqrt{6}+\sqrt{5}}{6-5}\\[5pt] &= \sqrt{6}+\sqrt{5}\,,\\[10pt] \frac{1}{\sqrt{7}-\sqrt{6}} &= \frac{1}{\sqrt{7}-\sqrt{6}}\cdot \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}\\[5pt] &= \frac{\sqrt{7}+\sqrt{6}}{(\sqrt{7})^{2}-(\sqrt{6})^{2}}\\[5pt] &= \frac{\sqrt{7}+\sqrt{6}}{7-6}\\[5pt] &= \sqrt{7}+\sqrt{6}\,\textrm{.} \end{align}

Now, we can subtract the terms and simplify the result,

\displaystyle \begin{align}

\frac{1}{\sqrt{6}-\sqrt{5}}-\frac{1}{\sqrt{7}-\sqrt{6}} &= \sqrt{6}+\sqrt{5}-(\sqrt{7}+\sqrt{6})\\[5pt] &= \sqrt{6}+\sqrt{5}-\sqrt{7}-\sqrt{6}\\[5pt] &= \sqrt{5}-\sqrt{7}\,\textrm{.} \end{align}

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