Svar 5.1:3

Förberedande kurs i matematik 1

(Skillnad mellan versioner)
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(Ny sida: {| width="100%" cellspacing="10px" |a) |width="100%" | \dfrac{x+1}{x^2-1} = \dfrac{1}{x-1} |- |b) |width="100%" | \left(\dfrac{5}{x}-1\right)(1-x) |- |c) |width="100%" | \dfrac{\frac{1}{2}}...)
Nuvarande version (3 juli 2009 kl. 13.28) (redigera) (ogör)
(\dfrac --> \displaystyle\frac)
 
Rad 1: Rad 1:
{| width="100%" cellspacing="10px"
{| width="100%" cellspacing="10px"
|a)
|a)
-
|width="100%" | \dfrac{x+1}{x^2-1} = \dfrac{1}{x-1}
+
|width="100%" | \displaystyle\frac{x+1}{x^2-1} = \displaystyle\frac{1}{x-1}
|-
|-
|b)
|b)
-
|width="100%" | \left(\dfrac{5}{x}-1\right)(1-x)
+
|width="100%" | \left(\displaystyle\frac{5}{x}-1\right)(1-x)
|-
|-
|c)
|c)
-
|width="100%" | \dfrac{\frac{1}{2}}{\frac{1}{3}+\frac{1}{4}}
+
|width="100%" | \displaystyle\frac{\frac{1}{2}}{\frac{1}{3}+\frac{1}{4}}
|-
|-
|d)
|d)
-
|width="100%" | \dfrac{1}{1+\dfrac{1}{1+x}}
+
|width="100%" | \displaystyle\frac{1}{1+\displaystyle\frac{1}{1+x}}
|}
|}

Nuvarande version

a) \displaystyle\frac{x+1}{x^2-1} = \displaystyle\frac{1}{x-1}
b) \left(\displaystyle\frac{5}{x}-1\right)(1-x)
c) \displaystyle\frac{\frac{1}{2}}{\frac{1}{3}+\frac{1}{4}}
d) \displaystyle\frac{1}{1+\displaystyle\frac{1}{1+x}}
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