1.4 Lösning 8b - Versionshistorik

Taifun den 3 augusti 2014 kl. 10.49

2014-08-03T10:49:46Z

Taifun den 3 augusti 2014 kl. 10.49

2014-08-03T10:49:25Z

Taifun den 5 mars 2011 kl. 16.09

2011-03-05T16:09:55Z

Taifun den 5 mars 2011 kl. 16.05

2011-03-05T16:05:44Z

Taifun: Created page with " ${1-x \over x+1} - {1+x \over 1-x} + {4\,x \over 1-x^2} \; = \; {1-x \over 1+x} - {1+x \over 1-x} + {4\,x \over (1+x)\cdot(1-x)} \; = \;$ $= \; {(1-x)\cdot(1..."$

2011-03-05T16:04:11Z

Created page with "<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> <math> = \; {(1-x)\cdot(1..."

Ny sida

← Äldre version		Versionen från 3 augusti 2014 kl. 10.49
Rad 6:		Rad 6:

	<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></big></big>		<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>

@@ 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 - 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>

← Äldre version		Versionen från 5 mars 2011 kl. 16.09
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)(1-x)} = {- 4x + 4x \over (1+x)(1-x)} = {0 \over ~~(1+x)(1-x)~~} = 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>

← Äldre version		Versionen från 5 mars 2011 kl. 16.05
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)} = {- 4x + 4x \over (1+x)~~\cdot~~(1-x)} = {0 \over (1+x)~~\cdot~~(1-x)} = 0 </math>	+	<math> = {1 - 2x + x^2 - 1 - 2x - x^2 + 4x \over (1+x)(1-x)} = {- 4x + 4x \over (1+x)(1-x)} = {0 \over (1+x)(1-x)} = 0 </math>

1.4 Lösning 8b - Versionshistorik

Taifun den 3 augusti 2014 kl. 10.49

Taifun den 3 augusti 2014 kl. 10.49

Taifun den 5 mars 2011 kl. 16.09

Taifun den 5 mars 2011 kl. 16.05

Taifun: Created page with " {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..."

Taifun: Created page with " ${1-x \over x+1} - {1+x \over 1-x} + {4\,x \over 1-x^2} \; = \; {1-x \over 1+x} - {1+x \over 1-x} + {4\,x \over (1+x)\cdot(1-x)} \; = \;$ $= \; {(1-x)\cdot(1..."$