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 | + | 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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+ | {{Displayed math||<math>\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} = \{\ y-x=-(x-y)\ \}\\[5pt] | ||
+ | &= \frac{-(x-y)^{2}}{y-x}+\frac{x^{2}}{y-x} = \frac{-(x-y)^{2}+x^{2}}{y-x}\\[5pt] | ||
+ | &= \frac{-(x^{2}-2xy+y^{2})+x^{2}}{y-x} = \frac{-x^{2}+2xy-y^{2}+x^{2}}{y-x}\\[5pt] | ||
+ | &= \frac{2xy-y^{2}}{y-x} = \frac{y(2x-y)}{y-x},\\[15pt] | ||
+ | \frac{y}{2x-y}-1 | ||
+ | &= \frac{y}{2x-y}-\frac{2x-y}{2x-y} = \frac{y-(2x-y)}{2x-y} = \frac{y-2x+y}{2x-y}\\[5pt] | ||
+ | & =\frac{2y-2x}{2x-y} = \frac{2(y-x)}{2x-y}\,\textrm{.} | ||
+ | \end{align}</math>}} | ||
Then, we multiply the factors together and simplify by elimination: | Then, we multiply the factors together and simplify by elimination: | ||
- | + | {{Displayed math||<math>\biggl(x-y+\frac{x^{2}}{y-x}\biggr) \biggl(\frac{y}{2x-y}-1\biggr) = \frac{y(2x-y)}{y-x}\cdot\frac{2(y-x)}{2x-y}=2y\,\textrm{.}</math>}} | |
- | <math>\ | + |
Version vom 11:24, 23. 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
Then, we multiply the factors together and simplify by elimination: