Lösung 3.1:6b

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The root sign in the denominator lies in a quadratic term and we therefore expand first the quadratic
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The root sign in the denominator lies in a quadratic term and we therefore expand first the quadratic
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<math>\left( \sqrt{3}-2 \right)^{2}</math>
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using the square rule
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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(\sqrt{3}-2)^2
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& \left( \sqrt{3}-2 \right)^{2}=\left( \sqrt{3} \right)^{2}-2\centerdot \sqrt{3}\centerdot 2+2^{2} \\
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&= (\sqrt{3}\,)^{2} - 2\cdot\sqrt{3}\cdot 2 + 2^{2}\\[5pt]
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& =3-4\sqrt{3}+4=7-4\sqrt{3} \\
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&= 3-4\sqrt{3}+4\\[5pt]
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\end{align}</math>
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&= 7-4\sqrt{3}\,\textrm{.}
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\end{align}</math>}}
Thus,
Thus,
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{{Displayed math||<math>\frac{1}{(\sqrt{3}-2)^{2}-2} = \frac{1}{7-4\sqrt{3}-2} = \frac{1}{5-4\sqrt{3}}</math>}}
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<math>\begin{align}
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and in the expression we can get rid of the root sign from the denominator by multiplying the top and bottom of the equation by the conjugate <math>5+4\sqrt{3}</math>,
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& \left( \sqrt{3}-2 \right)^{2}=\left( \sqrt{3} \right)^{2}-2\centerdot \sqrt{3}\centerdot 2+2^{2} \\
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& =3-4\sqrt{3}+4=7-4\sqrt{3} \\
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& \frac{1}{\left( \sqrt{3}-2 \right)^{2}-2}=\frac{1}{7-4\sqrt{3}-2}=\frac{1}{5-4\sqrt{3}} \\
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\end{align}</math>
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and in the expression we can get rid of the root sign from the denominator by multiplying the top and bottom of the equation by the conjugate
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<math>5+4\sqrt{3}</math>,
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<math>\begin{align}
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{{Displayed math||<math>\begin{align}
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& \frac{1}{5-4\sqrt{3}}=\frac{1}{5-4\sqrt{3}}\centerdot \frac{5+4\sqrt{3}}{5+4\sqrt{3}}=\frac{5+4\sqrt{3}}{5^{2}-\left( 4\sqrt{3} \right)^{2}} \\
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\frac{1}{5-4\sqrt{3}}
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& =\frac{5+4\sqrt{3}}{5^{2}-4^{2}\left( \sqrt{3} \right)^{2}}=\frac{5+4\sqrt{3}}{25-16\centerdot 3} \\
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&= \frac{1}{5-4\sqrt{3}}\cdot \frac{5+4\sqrt{3}}{5+4\sqrt{3}}\\[5pt]
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& =\frac{5+4\sqrt{3}}{-23}=-\frac{5+4\sqrt{3}}{23} \\
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&= \frac{5+4\sqrt{3}}{5^{2}-(4\sqrt{3})^{2}}\\[5pt]
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\end{align}</math>
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&= \frac{5+4\sqrt{3}}{5^{2}-4^{2}(\sqrt{3})^{2}}\\[5pt]
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&= \frac{5+4\sqrt{3}}{25-16\cdot 3}\\[5pt]
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&= \frac{5+4\sqrt{3}}{-23}\\[5pt]
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&= -\frac{5+4\sqrt{3}}{23}\,\textrm{.}
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\end{align}</math>}}

Version vom 11:53, 30. Sep. 2008

The root sign in the denominator lies in a quadratic term and we therefore expand first the quadratic

Vorlage:Displayed math

Thus,

Vorlage:Displayed math

and in the expression we can get rid of the root sign from the denominator by multiplying the top and bottom of the equation by the conjugate \displaystyle 5+4\sqrt{3},

Vorlage:Displayed math

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.1:6b“