1.4 Lösning 9c

\( \left(1 - {x^2 \over y^2}\right)\, \Bigg / \,\left(1 - {x \over y}\right) \, = \, \left({y^2 \over y^2} - {x^2 \over y^2}\right)\, \Bigg / \,\left({y \over y} - {x \over y}\right) \, = \)

\( = \, \left({y^2 - x^2 \over y^2}\right)\, \Bigg / \,\left({y - x \over y}\right) \, = \, \left({y^2 - x^2 \over y^2}\right)\, \cdot \,\left({y \over y - x}\right) \, = \)

\( = \, {(y+x)\,(y-x) \over y^2}\, \cdot \,{y \over y - x} \, = \, {(y+x)\,(y-x) \cdot y \over y^2 \cdot (y - x)} \, = \, = {y+x \over y} \)

\( \left({a^2 - 6\,a + 9 \over b^6}\right)\, \Bigg / \,\left({a - 3 \over b^5}\right) \, = \, \left({a^2 - 6\,a + 9 \over b^6}\right)\, \cdot \,\left({b^5 \over a - 3}\right) \, = \, \)

\( = \, {(a-3)^2 \over b^6}\, \cdot \,{b^5 \over a - 3} \, = \, {(a-3)^2 \cdot b^5 \over b^6 \cdot (a - 3)} \, = {a-3 \over b} \)