Skillnad mellan versioner av "1.4 Lösning 9a"
Från Mathonline
Taifun (Diskussion | bidrag) m |
Taifun (Diskussion | bidrag) |
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<big><big><math> \left({1 \over 2\,x - 1} + {1 \over 2\,x + 1}\right) \cdot {2\,x + 1 \over 2\,x} = {2\,x + 1 \over 2\,x\,(2\,x - 1)} + {1 \over 2\,x} = </math> | <big><big><math> \left({1 \over 2\,x - 1} + {1 \over 2\,x + 1}\right) \cdot {2\,x + 1 \over 2\,x} = {2\,x + 1 \over 2\,x\,(2\,x - 1)} + {1 \over 2\,x} = </math> | ||
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<math> = {2\,x + 1 \over 2\,x\,(2\,x - 1)} + {2\,x - 1 \over 2\,x\,(2\,x - 1)} = {2\,x + 1 + 2\,x - 1 \over 2\,x\,(2\,x - 1)} = </math> | <math> = {2\,x + 1 \over 2\,x\,(2\,x - 1)} + {2\,x - 1 \over 2\,x\,(2\,x - 1)} = {2\,x + 1 + 2\,x - 1 \over 2\,x\,(2\,x - 1)} = </math> | ||
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<math> = {4\,x \over 2\,x\,(2\,x - 1)} = {2 \over 2\,x - 1} </math></big></big> | <math> = {4\,x \over 2\,x\,(2\,x - 1)} = {2 \over 2\,x - 1} </math></big></big> | ||
Nuvarande version från 3 augusti 2014 kl. 23.11
\( \left({1 \over 2\,x - 1} + {1 \over 2\,x + 1}\right) \cdot {2\,x + 1 \over 2\,x} = {2\,x + 1 \over 2\,x\,(2\,x - 1)} + {1 \over 2\,x} = \)
\( = {2\,x + 1 \over 2\,x\,(2\,x - 1)} + {2\,x - 1 \over 2\,x\,(2\,x - 1)} = {2\,x + 1 + 2\,x - 1 \over 2\,x\,(2\,x - 1)} = \)
\( = {4\,x \over 2\,x\,(2\,x - 1)} = {2 \over 2\,x - 1} \)
