Lösung 2.1:3f

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Treating <math>4x</math> as one term, we can write
Treating <math>4x</math> as one term, we can write
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{{Displayed math||<math> \qquad 16x^2+8x+1=(4x)^2 +2\cdot 4x +1 </math>}}
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{{Abgesetzte Formel||<math> \qquad 16x^2+8x+1=(4x)^2 +2\cdot 4x +1 </math>}}
and since <math> y^2 +2y+1=(y+1)^2 </math> we obtain
and since <math> y^2 +2y+1=(y+1)^2 </math> we obtain
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{{Displayed math||<math> \qquad (4x)^2 +2\cdot 4x +1= (4x+1)^2 </math>.}}
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{{Abgesetzte Formel||<math> \qquad (4x)^2 +2\cdot 4x +1= (4x+1)^2 </math>.}}

Version vom 08:23, 22. Okt. 2008

Treating \displaystyle 4x as one term, we can write

\displaystyle \qquad 16x^2+8x+1=(4x)^2 +2\cdot 4x +1

and since \displaystyle y^2 +2y+1=(y+1)^2 we obtain

\displaystyle \qquad (4x)^2 +2\cdot 4x +1= (4x+1)^2 .
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.1:3f“