Skillnad mellan versioner av "1.4 Lösning 9c"
Från Mathonline
Taifun (Diskussion | bidrag) m |
Taifun (Diskussion | bidrag) m |
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| − | <big><big><math> = | + | <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} = {y \over y} + {x \over y} </math></big></big> <math> = 1 + </math> <big><big><math> {x \over y} </math></big></big> |
Nuvarande version från 3 augusti 2014 kl. 22.31
\( \left(1 - {x^2 \over y^2}\right)\, \Big / \,\left(1 - {x \over y}\right) \, = \, \left({y^2 \over y^2} - {x^2 \over y^2}\right)\, \Big / \,\left({y \over y} - {x \over y}\right) \, = \)
\( = \, \left({y^2 - x^2 \over y^2}\right)\, \Big / \,\left({y - x \over y}\right) \, = \, \left({y^2 - x^2 \over y^2}\right)\, \cdot \,\left({y \over y - x}\right) \, = \)
\( = {(y+x)\,(y-x) \over y^2} \cdot {y \over y - x} = {(y+x)\,(y-x) \cdot y \over y^2 \cdot (y - x)} = {y+x \over y} = {y \over y} + {x \over y} \) \( = 1 + \) \( {x \over y} \)
