Solution 3.1:7a

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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,
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<center> [[Image:3_1_7a.gif]] </center>
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<math>\begin{align}
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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{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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& \\
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\end{align}</math>
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Now, we can subtract the terms and simplify the result,
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<math>\begin{align}
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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}=\sqrt{5}-\sqrt{7}. \\
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\end{align}</math>

Revision as of 10:03, 23 September 2008

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}}\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}, \\ & \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}, \\ & \\ \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}-\left( \sqrt{7}+\sqrt{6} \right) \\ & =\sqrt{6}+\sqrt{5}-\sqrt{7}-\sqrt{6}=\sqrt{5}-\sqrt{7}. \\ \end{align}

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