Lösung 3.1:6b

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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The root sign in the denominator lies in a quadratic term and we therefore expand first the quadratic
The root sign in the denominator lies in a quadratic term and we therefore expand first the quadratic
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
(\sqrt{3}-2)^2
(\sqrt{3}-2)^2
&= (\sqrt{3}\,)^{2} - 2\cdot\sqrt{3}\cdot 2 + 2^{2}\\[5pt]
&= (\sqrt{3}\,)^{2} - 2\cdot\sqrt{3}\cdot 2 + 2^{2}\\[5pt]
Zeile 10: Zeile 10:
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>}}
+
{{Abgesetzte Formel||<math>\frac{1}{(\sqrt{3}-2)^{2}-2} = \frac{1}{7-4\sqrt{3}-2} = \frac{1}{5-4\sqrt{3}}</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 <math>5+4\sqrt{3}</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 <math>5+4\sqrt{3}</math>,
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\frac{1}{5-4\sqrt{3}}
\frac{1}{5-4\sqrt{3}}
&= \frac{1}{5-4\sqrt{3}}\cdot \frac{5+4\sqrt{3}}{5+4\sqrt{3}}\\[5pt]
&= \frac{1}{5-4\sqrt{3}}\cdot \frac{5+4\sqrt{3}}{5+4\sqrt{3}}\\[5pt]

Version vom 08:38, 22. Okt. 2008

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

\displaystyle \begin{align}

(\sqrt{3}-2)^2 &= (\sqrt{3}\,)^{2} - 2\cdot\sqrt{3}\cdot 2 + 2^{2}\\[5pt] &= 3-4\sqrt{3}+4\\[5pt] &= 7-4\sqrt{3}\,\textrm{.} \end{align}

Thus,

\displaystyle \frac{1}{(\sqrt{3}-2)^{2}-2} = \frac{1}{7-4\sqrt{3}-2} = \frac{1}{5-4\sqrt{3}}

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},

\displaystyle \begin{align}

\frac{1}{5-4\sqrt{3}} &= \frac{1}{5-4\sqrt{3}}\cdot \frac{5+4\sqrt{3}}{5+4\sqrt{3}}\\[5pt] &= \frac{5+4\sqrt{3}}{5^{2}-(4\sqrt{3})^{2}}\\[5pt] &= \frac{5+4\sqrt{3}}{5^{2}-4^{2}(\sqrt{3})^{2}}\\[5pt] &= \frac{5+4\sqrt{3}}{25-16\cdot 3}\\[5pt] &= \frac{5+4\sqrt{3}}{-23}\\[5pt] &= -\frac{5+4\sqrt{3}}{23}\,\textrm{.} \end{align}

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