Lösung 2.1:8c

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
Zeile 1: Zeile 1:
-
When we come across large and complicated expressions, we have to work step by step;
+
When we come across large and complicated expressions, we have to work step by step; as a first goal, we can multiply the top and bottom of the fraction
-
as a first goal, we can multiply the top and bottom of the fraction
+
{{Displayed math||<math>\frac{1}{1+\dfrac{1}{1+x}}</math>}}
 +
by <math>1+x</math>, so as to reduce it to an expression having one fraction sign
-
<math>\frac{1}{1+\frac{1}{1+x}}</math>
+
{{Displayed math||<math>\begin{align}
-
 
+
\frac{1}{1+\dfrac{1}{1+\dfrac{1}{1+x}}}
-
 
+
&= \frac{1}{1+\dfrac{1}{1+\dfrac{1}{1+x}}\cdot\dfrac{1+x}{1+x}}\\[8pt]
-
by
+
&= \frac{1}{1+\dfrac{1+x}{\Bigl(1+\dfrac{1}{1+x}\Bigr)(1+x)}}\\[8pt]
-
<math>1+x</math>, so as to reduce it to an expression having one fraction sign:
+
&= \frac{1}{1+\dfrac{1+x}{1+x+\dfrac{1+x}{1+x}}}\\[8pt]
-
 
+
&= \frac{1}{1+\dfrac{1+x}{1+x+1}}\\[8pt]
-
 
+
&= \frac{1}{1+\dfrac{x+1}{x+2}}\,\textrm{.}
-
<math>\begin{align}
+
\end{align}</math>}}
-
& \frac{1}{1+\frac{1}{1+\frac{1}{1+x}}}=\frac{1}{1+\frac{1}{1+\frac{1}{1+x}}\centerdot \frac{1+x}{1+x}}=\frac{1}{1+\frac{1+x}{\left( 1+\frac{1}{1+x} \right)\left( 1+x \right)}} \\
+
-
& \\
+
-
& =\frac{1}{1+\frac{1+x}{1+x+\frac{1+x}{1+x}}}=\frac{1}{1+\frac{1+x}{1+x+1}}=\frac{1}{1+\frac{x+1}{x+2}} \\
+
-
\end{align}</math>
+
-
 
+
The next step is to multiply the top and bottom of our new expression by
The next step is to multiply the top and bottom of our new expression by
-
<math>x+2</math>,
+
<math>x+2</math>, so as to obtain the final answer,
-
so as to obtain the final answer,
+
-
 
+
-
<math>\begin{align}
+
{{Displayed math||<math>\begin{align}
-
& \frac{1}{1+\frac{x+1}{x+2}}\centerdot \frac{x+2}{x+2}=\frac{x+2}{\left( 1+\frac{x+1}{x+2} \right)\left( x+2 \right)}=\frac{x+2}{x+2+\frac{x+1}{x+2}\left( x+2 \right)} \\
+
\frac{1}{1+\dfrac{x+1}{x+2}}\cdot\frac{x+2}{x+2}
-
& \\
+
&= \frac{x+2}{\Bigl(1+\dfrac{x+1}{x+2}\Bigr)(x+2)}\\[8pt]
-
& \frac{x+2}{x+2+x+1}=\frac{x+2}{2x+3} \\
+
&= \frac{x+2}{x+2+\dfrac{x+1}{x+2}(x+2)}\\[8pt]
-
& \\
+
&= \frac{x+2}{x+2+x+1}\\[8pt]
-
\end{align}</math>
+
&= \frac{x+2}{2x+3}\,\textrm{.}
 +
\end{align}</math>}}

Version vom 12:41, 23. Sep. 2008

When we come across large and complicated expressions, we have to work step by step; as a first goal, we can multiply the top and bottom of the fraction

Vorlage:Displayed math

by \displaystyle 1+x, so as to reduce it to an expression having one fraction sign

Vorlage:Displayed math

The next step is to multiply the top and bottom of our new expression by \displaystyle x+2, so as to obtain the final answer,

Vorlage:Displayed math