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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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
(5+4y)^2 &= 5^2+ 2\cdot 5 \cdot 4y +(4y)^2 \\[3pt]
(5+4y)^2 &= 5^2+ 2\cdot 5 \cdot 4y +(4y)^2 \\[3pt]
&= 25+10\cdot 4y + 4^2y^2\\[3pt]
&= 25+10\cdot 4y + 4^2y^2\\[3pt]

Version vom 08:21, 22. Okt. 2008

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

\displaystyle \begin{align}

(5+4y)^2 &= 5^2+ 2\cdot 5 \cdot 4y +(4y)^2 \\[3pt] &= 25+10\cdot 4y + 4^2y^2\\[3pt] &= 25+40y+16y^2\\[3pt] &= 16y^2 +40y + 25\,\textrm{.} \end{align}

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