Skillnad mellan versioner av "1.4 Lösning 9c"
Från Mathonline
Taifun (Diskussion | bidrag) m (Created page with "<math> \left(1 - {x^2 \over y^2}\right)\, \Bigg / \,\left(1 - {x \over y}\right) = </math>") |
Taifun (Diskussion | bidrag) m |
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| − | <math> \left(1 - {x^2 \over y^2}\right)\, \Bigg / \,\left(1 - {x \over y}\right) = </math> | + | <math> \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) \, = \, </math> |
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| + | <math> \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) \, = \, </math> | ||
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| + | <math> = \, {(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} </math> | ||
Versionen från 20 september 2012 kl. 23.26
\( \left(1 - {x^2 \over y^2}\right)\, \Bigg / \,\left(1 - {x \over y}\right) \, = \, \left({y^2 \over y^2} - {x^2 \over y^2}\right)\, \Bigg / \,\left({y \over y} - {x \over y}\right) \, = \, \)
\( \left({a^2 - 6\,a + 9 \over b^6}\right)\, \Bigg / \,\left({a - 3 \over b^5}\right) \, = \, \left({a^2 - 6\,a + 9 \over b^6}\right)\, \cdot \,\left({b^5 \over a - 3}\right) \, = \, \)
\( = \, {(a-3)^2 \over b^6}\, \cdot \,{b^5 \over a - 3} \, = \, {(a-3)^2 \cdot b^5 \over b^6 \cdot (a - 3)} \, = {a-3 \over b} \)
