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Lösung 3.4:1d

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
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K
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
We start by adding and taking away <math>x^2</math> in the numerator, so that, in combination with <math>x^3</math>, we obtain the expression <math>x^3+x^2 = x^2(x+1)</math> which can be simplified with the denominator <math>x+1</math>,
We start by adding and taking away <math>x^2</math> in the numerator, so that, in combination with <math>x^3</math>, we obtain the expression <math>x^3+x^2 = x^2(x+1)</math> which can be simplified with the denominator <math>x+1</math>,
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\frac{x^3+x+2}{x+1}
\frac{x^3+x+2}{x+1}
&= \frac{x^3+x^2-x^2+x+2}{x+1}\\[5pt]
&= \frac{x^3+x^2-x^2+x+2}{x+1}\\[5pt]
Zeile 13: Zeile 13:
<math>x+1</math>,
<math>x+1</math>,
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
x^2 + \frac{-x^2+x+2}{x+1}
x^2 + \frac{-x^2+x+2}{x+1}
&= x^2 + \frac{-x^2-x+x+x+2}{x+1}\\[5pt]
&= x^2 + \frac{-x^2-x+x+x+2}{x+1}\\[5pt]
Zeile 23: Zeile 23:
The last quotient divides perfectly and we obtain
The last quotient divides perfectly and we obtain
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{{Displayed math||<math>x^2-x+\frac{2x+2}{x+1}=x^2-x+2\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>x^2-x+\frac{2x+2}{x+1}=x^2-x+2\,\textrm{.}</math>}}
A quick check of whether
A quick check of whether
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{{Displayed math||<math>\frac{x^3+x+2}{x+1} = x^2-x+2\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>\frac{x^3+x+2}{x+1} = x^2-x+2\,\textrm{.}</math>}}
is the correct answer is to investigate whether
is the correct answer is to investigate whether
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{{Displayed math||<math>x^3+x+2 = (x^2-x+2)(x+1)</math>}}
+
{{Abgesetzte Formel||<math>x^3+x+2 = (x^2-x+2)(x+1)</math>}}
holds. If we expand the right-hand side, we see that the relation really does hold
holds. If we expand the right-hand side, we see that the relation really does hold
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
(x^2-x+2)(x+1) = x^3+x^2-x^2-x+2x+2 = x^3+x+2\,\textrm{.}
(x^2-x+2)(x+1) = x^3+x^2-x^2-x+2x+2 = x^3+x+2\,\textrm{.}
\end{align}</math>}}
\end{align}</math>}}

Version vom 13:15, 10. Mär. 2009

We start by adding and taking away x2 in the numerator, so that, in combination with x3, we obtain the expression x3+x2=x2(x+1) which can be simplified with the denominator x+1,

x+1x3+x+2=x+1x3+x2−x2+x+2=x+1x3+x2+x+1−x2+x+2=x+1x2(x+1)+x+1−x2+x+2=x2+x+1−x2+x+2.

The term −x2 in the remaining quotient needs to complemented with −x so that we get −x2−x=−x(x+1), which is divisible by x+1,

x2+x+1−x2+x+2=x2+x+1−x2−x+x+x+2=x2+x+1−x2−x+x+12x+2=x2+x+1−x(x+1)+x+12x+2=x2−x+x+12x+2.

The last quotient divides perfectly and we obtain

x2−x+x+12x+2=x2−x+2.

A quick check of whether

x+1x3+x+2=x2−x+2.

is the correct answer is to investigate whether

x3+x+2=(x2−x+2)(x+1)

holds. If we expand the right-hand side, we see that the relation really does hold

(x2−x+2)(x+1)=x3+x2−x2−x+2x+2=x3+x+2.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.4:1d“