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		<title>1.4 Lösning 9c - Versionshistorik</title>
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		<updated>2026-11-05T14:19:44Z</updated>
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		<title>Taifun den 3 augusti 2014 kl. 22.31</title>
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				<updated>2014-08-03T22:31:16Z</updated>
		
		<summary type="html">&lt;p&gt;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;Ny sida&lt;/b&gt;&lt;/p&gt;&lt;div&gt;&amp;lt;big&amp;gt;&amp;lt;math&amp;gt; \left(1 - {x^2 \over y^2}\right)\, \Big / \,\left(1 - {x \over y}\right) \, = \, \left({y^2 \over y^2} - {x^2 \over y^2}\right)\, \Big / \,\left({y \over y} - {x \over y}\right) \, = &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; = \, \left({y^2 - x^2 \over y^2}\right)\, \Big / \,\left({y - x \over y}\right) \, = \, \left({y^2 - x^2 \over y^2}\right)\, \cdot \,\left({y \over y - x}\right) \, = &amp;lt;/math&amp;gt;&amp;lt;/big&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;big&amp;gt;&amp;lt;big&amp;gt;&amp;lt;math&amp;gt; = {(y+x)\,(y-x) \over y^2} \cdot {y \over y - x} = {(y+x)\,(y-x) \cdot y \over y^2 \cdot (y - x)} = {y+x \over y} = {y \over y} + {x \over y} &amp;lt;/math&amp;gt;&amp;lt;/big&amp;gt;&amp;lt;/big&amp;gt; &amp;lt;math&amp;gt; = 1 + &amp;lt;/math&amp;gt; &amp;lt;big&amp;gt;&amp;lt;big&amp;gt;&amp;lt;math&amp;gt; {x \over y} &amp;lt;/math&amp;gt;&amp;lt;/big&amp;gt;&amp;lt;/big&amp;gt;&lt;/div&gt;</summary>
		<author><name>Taifun</name></author>	</entry>

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