Aus Online Mathematik Brückenkurs 1

Version vom 08:46, 9. Sep. 2008 (bearbeiten) Tekbot (Diskussion | Beiträge) K (Lösning 2.2:3d moved to Solution 2.2:3d: Robot: moved page) ← Zum vorherigen Versionsunterschied		Version vom 15:48, 17. Sep. 2008 (bearbeiten) (rückgängig) Ian (Diskussion | Beiträge) Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
-	{{NAVCONTENT_START}}	+	(The exercise is taken from an actual exam in Spring Term 1945!)
-	<center> [[Image:2_2_3d-1(3).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	There are no common factors on the left-hand side which we can take out, so we choose to expand the three terms on the left-hand side:
-	{{NAVCONTENT_START}}	+
-	<center> [[Image:2_2_3d-2(3).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	<math>\begin{align}
-	{{NAVCONTENT_START}}	+	& \left( \frac{2}{x}-3 \right)\left( \frac{1}{4x}+\frac{1}{2} \right)=\frac{2}{x}\centerdot \frac{1}{4x}-\frac{2}{x}\centerdot \frac{1}{2}-3\centerdot \frac{1}{4x}-3\centerdot \frac{1}{2} \\
-	<center> [[Image:2_2_3d-3(3).gif]] </center>	+	& =\frac{1}{2x^{2}}+\frac{1}{x}-\frac{3}{4x}-\frac{3}{2}=\frac{1}{2x^{2}}+\frac{1}{4x}-\frac{3}{2}, \\
-	{{NAVCONTENT_STOP}}	+	& \\
		+	& \left( \frac{1}{2x}-\frac{2}{3} \right)^{2}=\frac{1}{\left( 2x \right)^{2}}-2\centerdot \frac{1}{2x}\centerdot \frac{2}{3}+\left( \frac{2}{3} \right)^{2} \\
		+	& =\frac{1}{4x^{2}}-\frac{2}{3x}+\frac{4}{9}, \\
		+	& \\
		+	& \\
		+	\end{align}</math>
		+
		+
		+	<math>\begin{align}
		+	& \left( \frac{1}{2x}+\frac{1}{3} \right)\left( \frac{1}{2x}-\frac{1}{3} \right)=\left\{ \text{conjugate rule} \right\}=\frac{1}{\left( 2x \right)^{2}}-\frac{1}{3^{2}} \\
		+	& =\frac{1}{4x^{2}}-\frac{1}{9} \\
		+	\end{align}</math>
		+
		+
		+
		+	Collecting up terms, the left-hand side becomes
		+
		+
		+	<math>\begin{align}
		+	& \left( \frac{1}{2x^{2}}+\frac{1}{4x}-\frac{3}{2} \right)-\left( \frac{1}{4x^{2}}-\frac{2}{3x}+\frac{4}{9} \right)-\left( \frac{1}{4x^{2}}-\frac{1}{9} \right) \\
		+	& =\left( \frac{1}{2}-\frac{1}{4}-\frac{1}{4} \right)\frac{1}{x^{2}}+\left( \frac{1}{4}+\frac{2}{3} \right)\frac{1}{x}+\left( -\frac{3}{2}-\frac{4}{9}+\frac{1}{9} \right) \\
		+	& =\frac{2-1-1}{4}\frac{1}{x^{2}}+\frac{3+2\centerdot 4}{3\centerdot 4}\frac{1}{x}+\frac{-3\centerdot 9-4\centerdot 2+1\centerdot 2}{2\centerdot 9} \\
		+	& =\frac{11}{3\centerdot 4}\centerdot \frac{1}{x}-\frac{33}{2\centerdot 9} \\
		+	\end{align}</math>
		+
		+
		+
		+	and because
		+	<math>33=3\centerdot 11</math>,
		+	<math>9=3\centerdot 3</math>
		+	and
		+	<math>4=2\centerdot 2</math>, the whole equation can rewritten as
		+
		+
		+	<math>\frac{11}{3\centerdot 2\centerdot 2}\centerdot \frac{1}{x}-\frac{3\centerdot 11}{2\centerdot 3\centerdot 3}=0</math>
		+
		+
		+	Taking out common factors, we get
		+
		+
		+	<math>\frac{11}{3\centerdot 2}\left( \frac{1}{2x}-1 \right)=0</math>
		+
		+
		+	and then we see that the equation has the solution
		+	<math>x={1}/{2}\;</math>.
		+
		+	Finally, we substitute
		+	<math>x={1}/{2}\;</math>
		+	into the original equation to check that we have calculated correctly.
		+
		+
		+	<math>\begin{align}
		+	& \left( \frac{2}{\frac{1}{2}}-3 \right)\left( \frac{1}{4\centerdot \frac{1}{2}}+\frac{1}{2} \right)-\left( \frac{1}{2\centerdot \frac{1}{2}}-\frac{2}{3} \right)^{2}-\left( \frac{1}{2\centerdot \frac{1}{2}}+\frac{1}{3} \right)\left( \frac{1}{2\centerdot \frac{1}{2}}-\frac{1}{3} \right) \\
		+	& \\
		+	& =\left( 4-3 \right)\left( \frac{1}{2}+\frac{1}{2} \right)-\left( 1-\frac{2}{3} \right)^{2}-\left( 1+\frac{1}{3} \right)\left( 1-\frac{1}{3} \right) \\
		+	& \\
		+	& =1-\left( \frac{1}{3} \right)^{2}-\frac{4}{3}\centerdot \frac{2}{3}=1-\frac{1}{9}-\frac{8}{9}=0 \\
		+	\end{align}</math>

---

Version vom 15:48, 17. Sep. 2008

(The exercise is taken from an actual exam in Spring Term 1945!)

There are no common factors on the left-hand side which we can take out, so we choose to expand the three terms on the left-hand side:

\displaystyle \begin{align} & \left( \frac{2}{x}-3 \right)\left( \frac{1}{4x}+\frac{1}{2} \right)=\frac{2}{x}\centerdot \frac{1}{4x}-\frac{2}{x}\centerdot \frac{1}{2}-3\centerdot \frac{1}{4x}-3\centerdot \frac{1}{2} \\ & =\frac{1}{2x^{2}}+\frac{1}{x}-\frac{3}{4x}-\frac{3}{2}=\frac{1}{2x^{2}}+\frac{1}{4x}-\frac{3}{2}, \\ & \\ & \left( \frac{1}{2x}-\frac{2}{3} \right)^{2}=\frac{1}{\left( 2x \right)^{2}}-2\centerdot \frac{1}{2x}\centerdot \frac{2}{3}+\left( \frac{2}{3} \right)^{2} \\ & =\frac{1}{4x^{2}}-\frac{2}{3x}+\frac{4}{9}, \\ & \\ & \\ \end{align}

\displaystyle \begin{align} & \left( \frac{1}{2x}+\frac{1}{3} \right)\left( \frac{1}{2x}-\frac{1}{3} \right)=\left\{ \text{conjugate rule} \right\}=\frac{1}{\left( 2x \right)^{2}}-\frac{1}{3^{2}} \\ & =\frac{1}{4x^{2}}-\frac{1}{9} \\ \end{align}

Collecting up terms, the left-hand side becomes

\displaystyle \begin{align} & \left( \frac{1}{2x^{2}}+\frac{1}{4x}-\frac{3}{2} \right)-\left( \frac{1}{4x^{2}}-\frac{2}{3x}+\frac{4}{9} \right)-\left( \frac{1}{4x^{2}}-\frac{1}{9} \right) \\ & =\left( \frac{1}{2}-\frac{1}{4}-\frac{1}{4} \right)\frac{1}{x^{2}}+\left( \frac{1}{4}+\frac{2}{3} \right)\frac{1}{x}+\left( -\frac{3}{2}-\frac{4}{9}+\frac{1}{9} \right) \\ & =\frac{2-1-1}{4}\frac{1}{x^{2}}+\frac{3+2\centerdot 4}{3\centerdot 4}\frac{1}{x}+\frac{-3\centerdot 9-4\centerdot 2+1\centerdot 2}{2\centerdot 9} \\ & =\frac{11}{3\centerdot 4}\centerdot \frac{1}{x}-\frac{33}{2\centerdot 9} \\ \end{align}

and because \displaystyle 33=3\centerdot 11, \displaystyle 9=3\centerdot 3 and \displaystyle 4=2\centerdot 2, the whole equation can rewritten as

\displaystyle \frac{11}{3\centerdot 2\centerdot 2}\centerdot \frac{1}{x}-\frac{3\centerdot 11}{2\centerdot 3\centerdot 3}=0

Taking out common factors, we get

\displaystyle \frac{11}{3\centerdot 2}\left( \frac{1}{2x}-1 \right)=0

and then we see that the equation has the solution \displaystyle x={1}/{2}\;.

Finally, we substitute \displaystyle x={1}/{2}\; into the original equation to check that we have calculated correctly.

\displaystyle \begin{align} & \left( \frac{2}{\frac{1}{2}}-3 \right)\left( \frac{1}{4\centerdot \frac{1}{2}}+\frac{1}{2} \right)-\left( \frac{1}{2\centerdot \frac{1}{2}}-\frac{2}{3} \right)^{2}-\left( \frac{1}{2\centerdot \frac{1}{2}}+\frac{1}{3} \right)\left( \frac{1}{2\centerdot \frac{1}{2}}-\frac{1}{3} \right) \\ & \\ & =\left( 4-3 \right)\left( \frac{1}{2}+\frac{1}{2} \right)-\left( 1-\frac{2}{3} \right)^{2}-\left( 1+\frac{1}{3} \right)\left( 1-\frac{1}{3} \right) \\ & \\ & =1-\left( \frac{1}{3} \right)^{2}-\frac{4}{3}\centerdot \frac{2}{3}=1-\frac{1}{9}-\frac{8}{9}=0 \\ \end{align}