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Lösung 2.1:1g

Aus Online Mathematik Brückenkurs 1

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The expression in the exercise is of the form <math> (a-b)^2 </math>, where <math> a=y^2</math> and <math> b=3x^2 </math>. With the help of the squaring rule <math> (a-b)^2 =a^2 -2ab +b^2 </math>, we have
The expression in the exercise is of the form <math> (a-b)^2 </math>, where <math> a=y^2</math> and <math> b=3x^2 </math>. With the help of the squaring rule <math> (a-b)^2 =a^2 -2ab +b^2 </math>, we have
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
(y^2-3x^3)^2 &= (y^2)^2 -2\cdot y^2\cdot 3x^3 +(3x^3)^2 \\[3pt]
(y^2-3x^3)^2 &= (y^2)^2 -2\cdot y^2\cdot 3x^3 +(3x^3)^2 \\[3pt]
&= y^{2\cdot 2} -6x^3y^2 +3^2x^{3\cdot 2}\\[3pt]
&= y^{2\cdot 2} -6x^3y^2 +3^2x^{3\cdot 2}\\[3pt]

Version vom 08:21, 22. Okt. 2008

The expression in the exercise is of the form (a−b)2, where a=y2 and b=3x2. With the help of the squaring rule (a−b)2=a2−2ab+b2, we have

(y2−3x3)2=(y2)2−2y23x3+(3x3)2=y22−6x3y2+32x32=y4−6x3y2+9x6=9x6−6x3y2+y4.
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.1:1g“