Skillnad mellan versioner av "1.4 Lösning 9c"
Från Mathonline
Taifun (Diskussion | bidrag) m |
Taifun (Diskussion | bidrag) m |
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| Rad 1: | Rad 1: | ||
| − | + | <big><math> \left(1 - {x^2 \over y^2}\right)\, \Big / \,\left(1 - {x \over y}\right) \, = \, \left({y^2 \over y^2} - {x^2 \over y^2}\right)\, \Big / \,\left({y \over y} - {x \over y}\right) \, = </math> | |
| − | <math> = \, \left({y^2 - x^2 \over y^2}\right)\, \Big / \,\left({y - x \over y}\right) \, = \, \left({y^2 - x^2 \over y^2}\right)\, \cdot \,\left({y \over y - x}\right) \, = </math> | + | <math> = \, \left({y^2 - x^2 \over y^2}\right)\, \Big / \,\left({y - x \over y}\right) \, = \, \left({y^2 - x^2 \over y^2}\right)\, \cdot \,\left({y \over y - x}\right) \, = </math></big> |
| − | <math> = \, {(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} \, = \, 1 + {x \over y} </math></big></big> | + | <big><big><math> = \, {(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} \, = \, 1 + {x \over y} </math></big></big> |
Versionen från 3 augusti 2014 kl. 23.22
\( \left(1 - {x^2 \over y^2}\right)\, \Big / \,\left(1 - {x \over y}\right) \, = \, \left({y^2 \over y^2} - {x^2 \over y^2}\right)\, \Big / \,\left({y \over y} - {x \over y}\right) \, = \)
\( = \, \left({y^2 - x^2 \over y^2}\right)\, \Big / \,\left({y - x \over y}\right) \, = \, \left({y^2 - x^2 \over y^2}\right)\, \cdot \,\left({y \over y - x}\right) \, = \)
\( = \, {(y+x)\,(y-x) \over y^2}\, \cdot \,{y \over y - x} \, = \, {(y+x)\,(y-x) \cdot y \over y^2 \cdot (y - x)} \, = \, {y+x \over y} \, = \, 1 + {x \over y} \)
