Lösung 2.1:6a

Aus Online Mathematik Brückenkurs 1

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Before we try dealing with the whole expression, we focus on simplifying the two factors individually by rewriting them using a common denominator:
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<center> [[Image:2_1_6a.gif]] </center>
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<math>\begin{align}
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& x-y+\frac{x^{2}}{y-x}=\frac{\left( x-y \right)\left( y-x \right)}{y-x}+\frac{x^{2}}{y-x} \\
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& \left\{ y-x=-\left( x-y \right) \right\} \\
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& =\frac{-\left( x-y \right)^{2}}{y-x}+\frac{x^{2}}{y-x}=\frac{-\left( x-y \right)^{2}+x^{2}}{y-x}=\frac{-\left( x^{2}-2xy+y^{2} \right)+x^{2}}{y-x} \\
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& =\frac{-x^{2}+2xy-y^{2}+x^{2}}{y-x}=\frac{2xy-y^{2}}{y-x}=\frac{y\left( 2x-y \right)}{y-x}, \\
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\end{align}</math>
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<math>\begin{align}
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& \frac{y}{2x-y}-1=\frac{y}{2x-y}-\frac{2x-y}{2x-y}=\frac{y-\left( 2x-y \right)}{2x-y}=\frac{y-2x+y}{2x-y} \\
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& =\frac{2y-2x}{2x-y}=\frac{2\left( y-x \right)}{2x-y} \\
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\\
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\end{align}</math>
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Then, we multiply the factors together and simplify by elimination:
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<math>\left( x-y+\frac{x^{2}}{y-x} \right)\left( \frac{y}{2x-y}-1 \right)=\frac{y\left( 2x-y \right)}{y-x}\centerdot \frac{2\left( y-x \right)}{2x-y}=2y.</math>

Version vom 10:37, 16. Sep. 2008

Before we try dealing with the whole expression, we focus on simplifying the two factors individually by rewriting them using a common denominator:


\displaystyle \begin{align} & x-y+\frac{x^{2}}{y-x}=\frac{\left( x-y \right)\left( y-x \right)}{y-x}+\frac{x^{2}}{y-x} \\ & \left\{ y-x=-\left( x-y \right) \right\} \\ & =\frac{-\left( x-y \right)^{2}}{y-x}+\frac{x^{2}}{y-x}=\frac{-\left( x-y \right)^{2}+x^{2}}{y-x}=\frac{-\left( x^{2}-2xy+y^{2} \right)+x^{2}}{y-x} \\ & =\frac{-x^{2}+2xy-y^{2}+x^{2}}{y-x}=\frac{2xy-y^{2}}{y-x}=\frac{y\left( 2x-y \right)}{y-x}, \\ \end{align}


\displaystyle \begin{align} & \frac{y}{2x-y}-1=\frac{y}{2x-y}-\frac{2x-y}{2x-y}=\frac{y-\left( 2x-y \right)}{2x-y}=\frac{y-2x+y}{2x-y} \\ & =\frac{2y-2x}{2x-y}=\frac{2\left( y-x \right)}{2x-y} \\ \\ \end{align}


Then, we multiply the factors together and simplify by elimination:


\displaystyle \left( x-y+\frac{x^{2}}{y-x} \right)\left( \frac{y}{2x-y}-1 \right)=\frac{y\left( 2x-y \right)}{y-x}\centerdot \frac{2\left( y-x \right)}{2x-y}=2y.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.1:6a“