Solution 2.1:1d
From Förberedande kurs i matematik 1
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} |
