Solution 2.1:1d
From Förberedande kurs i matematik 1
(Difference between revisions)
(Ny sida: {{NAVCONTENT_START}} <center> Bild:2_1_1d.gif </center> {{NAVCONTENT_STOP}}) |
m |
||
| (3 intermediate revisions not shown.) | |||
| Line 1: | Line 1: | ||
| - | {{ | + | After <math> x^3y^2 </math> are multiplied inside the bracket, we can eliminate factors which occur in both the numerator and denominator, |
| - | < | + | |
| - | {{ | + | {{Displayed math||<math>\begin{align} |
| + | x^3y^2\Big( \frac{1}{y} - \frac{1}{xy} +1 \Big) &= x^3y^2 \cdot\frac{1}{y} -x^3y^2 \cdot \frac{1}{xy} +x^3y^2\cdot 1 \\ | ||
| + | &=\frac{x^3y^2}{y} -\frac{x^3y^2}{xy} +x^3y^2 \\ | ||
| + | &=x^3y - x^2y +x^3y^2\,, | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | where we have used | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | \frac{x^3y^2}{y} &= \frac{x^3\cdot y\cdot{}\rlap{/}y}{\rlap{/}y}= x^3y\,,\\[5pt] | ||
| + | \frac{x^3y^2}{xy} &= \frac{\rlap{/}x\cdot x\cdot x \cdot y \cdot {}\rlap{/}y}{\rlap{/}x\cdot {}\rlap{/}y} = x\cdot x\cdot y = x^2y\,\textrm{.}\end{align}</math>}} | ||
Current revision
After \displaystyle x^3y^2 are multiplied inside the bracket, we can eliminate factors which occur in both the numerator and denominator,
| \displaystyle \begin{align}
x^3y^2\Big( \frac{1}{y} - \frac{1}{xy} +1 \Big) &= x^3y^2 \cdot\frac{1}{y} -x^3y^2 \cdot \frac{1}{xy} +x^3y^2\cdot 1 \\ &=\frac{x^3y^2}{y} -\frac{x^3y^2}{xy} +x^3y^2 \\ &=x^3y - x^2y +x^3y^2\,, \end{align} |
where we have used
| \displaystyle \begin{align}
\frac{x^3y^2}{y} &= \frac{x^3\cdot y\cdot{}\rlap{/}y}{\rlap{/}y}= x^3y\,,\\[5pt] \frac{x^3y^2}{xy} &= \frac{\rlap{/}x\cdot x\cdot x \cdot y \cdot {}\rlap{/}y}{\rlap{/}x\cdot {}\rlap{/}y} = x\cdot x\cdot y = x^2y\,\textrm{.}\end{align} |
