Skillnad mellan versioner av "1.4 Lösning 8b"
Från Mathonline
Taifun (Diskussion | bidrag) m |
Taifun (Diskussion | bidrag) m |
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| Rad 1: | Rad 1: | ||
| − | <math> {1-x \over x+1} - {1+x \over 1-x} + {4\,x \over 1-x^2} \; = \; {1-x \over 1+x} - {1+x \over 1-x} + {4\,x \over (1+x)\cdot(1-x)} \; = \; </math> | + | <big><big><math> {1-x \over x+1} - {1+x \over 1-x} + {4\,x \over 1-x^2} \; = \; {1-x \over 1+x} - {1+x \over 1-x} + {4\,x \over (1+x)\cdot(1-x)} \; = \; </math> |
| Rad 7: | Rad 7: | ||
<math> = {(1-x)^2 - (1+x)^2 + 4\,x \over (1+x)\cdot(1-x)} = {1-2\,x+x^2 - (1+2\,x+x^2) + 4\,x \over (1+x)\cdot(1-x)} = </math> | <math> = {(1-x)^2 - (1+x)^2 + 4\,x \over (1+x)\cdot(1-x)} = {1-2\,x+x^2 - (1+2\,x+x^2) + 4\,x \over (1+x)\cdot(1-x)} = </math> | ||
| − | <math> = {1 - 2x + x^2 - 1 - 2x - x^2 + 4x \over (1+x)\cdot(1-x)} \; = \; {- 4\,x + 4\,x \over (1+x)\cdot(1-x)} \; = \; {0 \over \dots} \; = \; 0 </math> | + | <math> = {1 - 2x + x^2 - 1 - 2x - x^2 + 4x \over (1+x)\cdot(1-x)} \; = \; {- 4\,x + 4\,x \over (1+x)\cdot(1-x)} \; = \; {0 \over \dots} \; = \; 0 </math></big></big> |
Versionen från 3 augusti 2014 kl. 11.49
\( {1-x \over x+1} - {1+x \over 1-x} + {4\,x \over 1-x^2} \; = \; {1-x \over 1+x} - {1+x \over 1-x} + {4\,x \over (1+x)\cdot(1-x)} \; = \; \)
\( = \; {(1-x)\cdot(1-x) \over (1+x)\cdot(1-x)} - {(1+x)\cdot(1+x) \over (1-x)\cdot(1+x)} + {4\,x \over (1+x)\cdot(1-x)} \; = \; \)
\( = {(1-x)^2 - (1+x)^2 + 4\,x \over (1+x)\cdot(1-x)} = {1-2\,x+x^2 - (1+2\,x+x^2) + 4\,x \over (1+x)\cdot(1-x)} = \)
\( = {1 - 2x + x^2 - 1 - 2x - x^2 + 4x \over (1+x)\cdot(1-x)} \; = \; {- 4\,x + 4\,x \over (1+x)\cdot(1-x)} \; = \; {0 \over \dots} \; = \; 0 \)
