Skillnad mellan versioner av "1.4 Lösning 9c"

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<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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<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)\, \Bigg / \,\left({y - x \over y}\right) \, = \, \left({y^2 - x^2 \over y^2}\right)\, \cdot \,\left({y \over y - x}\right) \, = </math>
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<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> \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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<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>
 
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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>
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Nuvarande version från 3 augusti 2014 kl. 23.31

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


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


\( = {(y+x)\,(y-x) \over y^2} \cdot {y \over y - x} = {(y+x)\,(y-x) \cdot y \over y^2 \cdot (y - x)} = {y+x \over y} = {y \over y} + {x \over y} \) \( = 1 + \) \( {x \over y} \)