Lösung 2.1:1f

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The squaring rule <math> (a+b)^2 = a^2+2ab+b^2</math> with <math>a=5 </math> and <math> b=4y </math> gives
The squaring rule <math> (a+b)^2 = a^2+2ab+b^2</math> with <math>a=5 </math> and <math> b=4y </math> gives
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<math>\qquad \begin{align} (5+4y)^2 &= 5^2+ 2\cdot 5 \cdot 4y +(4y)^2 \\
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{{Displayed math||<math>\begin{align}
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&= 25+10\cdot 4y + 4^2y^2\\
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(5+4y)^2 &= 5^2+ 2\cdot 5 \cdot 4y +(4y)^2 \\[3pt]
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&= 25+40y+16y^4\\
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&= 25+10\cdot 4y + 4^2y^2\\[3pt]
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&= 16y^2 +40y + 25
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&= 25+40y+16y^2\\[3pt]
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\end{align}
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&= 16y^2 +40y + 25\,\textrm{.}
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</math>
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\end{align}</math>}}
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Version vom 07:56, 23. Sep. 2008

The squaring rule \displaystyle (a+b)^2 = a^2+2ab+b^2 with \displaystyle a=5 and \displaystyle b=4y gives

Vorlage:Displayed math

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.1:1f“