Lösung 2.1:1e

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If we use the rule for squaring <math>(a-b)^2 = a^2-2ab+b^2 </math> with <math> a=x </math> and <math> b=7</math>, we obtain directly that
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If we use the rule for squaring <math>(a-b)^2 = a^2-2ab+b^2 </math> with <math> a=x </math> and <math> b=7, </math> we obtain directly that
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:<math> (x-7)^2=x^2-2 \cdot x \cdot 7 + 7^2 = x^2-14x+49.</math>
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{{Displayed math||<math> (x-7)^2=x^2-2 \cdot x \cdot 7 + 7^2 = x^2-14x+49\,\textrm{.}</math>}}
An alternative is to write the square as <math> (x-7)\cdot (x-7)</math> and then multiply the brackets in two steps
An alternative is to write the square as <math> (x-7)\cdot (x-7)</math> and then multiply the brackets in two steps
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<math>
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{{Displayed math||<math>\begin{align}
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\qquad
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(x-7)\cdot (x-7) &= (x-7)\cdot x - (x-7)\cdot 7 \\[3pt]
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\begin{align}
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&= x\cdot x-7 \cdot x -(x\cdot 7 - 7\cdot 7) \\[3pt]
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(x-7)\cdot (x-7) &= (x-7)\cdot x - (x-7)\cdot 7 \\
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&= x^2 -7x-(7x-49)\\[3pt]
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&= x\cdot x-7 \cdot x -(x\cdot 7 - 7\cdot 7) \\
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& \stackrel{*}= x^2-7x-7x+49 \\[3pt]
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&= x^2 -7x-(7x-49)\\
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&= x^2-(7+7)x+49\\[3pt]
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& \stackrel{*}= x^2-7x-7x+49 \\
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&= x^2-14x+49\,\textrm{.}
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&= x^2-(7+7)x+49\\
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\end{align}</math>}}
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&= x^2-14x+49
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\end{align}
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</math>
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In the line that has been marked with an asterisk, we have removed the bracket and at the same time changed signs on all terms inside the bracket.
In the line that has been marked with an asterisk, we have removed the bracket and at the same time changed signs on all terms inside the bracket.
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Version vom 07:53, 23. Sep. 2008

If we use the rule for squaring \displaystyle (a-b)^2 = a^2-2ab+b^2 with \displaystyle a=x and \displaystyle b=7, we obtain directly that

Vorlage:Displayed math

An alternative is to write the square as \displaystyle (x-7)\cdot (x-7) and then multiply the brackets in two steps

Vorlage:Displayed math

In the line that has been marked with an asterisk, we have removed the bracket and at the same time changed signs on all terms inside the bracket.

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