Lösung 2.1:1e
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
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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 | 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 | ||
| - | {{ | + | {{Abgesetzte Formel||<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 | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
(x-7)\cdot (x-7) &= (x-7)\cdot x - (x-7)\cdot 7 \\[3pt] | (x-7)\cdot (x-7) &= (x-7)\cdot x - (x-7)\cdot 7 \\[3pt] | ||
&= x\cdot x-7 \cdot x -(x\cdot 7 - 7\cdot 7) \\[3pt] | &= x\cdot x-7 \cdot x -(x\cdot 7 - 7\cdot 7) \\[3pt] | ||
Version vom 08:20, 22. Okt. 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
| \displaystyle (x-7)^2=x^2-2 \cdot x \cdot 7 + 7^2 = x^2-14x+49\,\textrm{.} |
An alternative is to write the square as \displaystyle (x-7)\cdot (x-7) and then multiply the brackets in two steps
| \displaystyle \begin{align}
(x-7)\cdot (x-7) &= (x-7)\cdot x - (x-7)\cdot 7 \\[3pt] &= x\cdot x-7 \cdot x -(x\cdot 7 - 7\cdot 7) \\[3pt] &= x^2 -7x-(7x-49)\\[3pt] & \stackrel{*}= x^2-7x-7x+49 \\[3pt] &= x^2-(7+7)x+49\\[3pt] &= x^2-14x+49\,\textrm{.} \end{align} |
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.
