Svar 1.2:1

Förberedande kurs i matematik 2

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(Ny sida: {| width="100%" cellspacing="10px" |a) |width="33%"| <math>\cos^2x-\sin^2x=\cos2x</math> |b) |width="33%"| <math>2x\ln x+ x</math> |c) |width="33%"| <math>\displaystyle\frac{x^2+2x-1}{(x+1)...)
Nuvarande version (4 april 2008 kl. 07.58) (redigera) (ogör)
(Ny sida: {| width="100%" cellspacing="10px" |a) |width="33%"| <math>\cos^2x-\sin^2x=\cos2x</math> |b) |width="33%"| <math>2x\ln x+ x</math> |c) |width="33%"| <math>\displaystyle\frac{x^2+2x-1}{(x+1)...)
 

Nuvarande version

a) \displaystyle \cos^2x-\sin^2x=\cos2x b) \displaystyle 2x\ln x+ x c) \displaystyle \displaystyle\frac{x^2+2x-1}{(x+1)^2}=1-\frac{2}{(x+1)^2}
d) \displaystyle \displaystyle\frac{\cos x}{x}-\frac{\sin x}{x^2} e) \displaystyle \displaystyle\frac{1}{\ln x}-\frac{1}{(\ln x)^2} f) \displaystyle \displaystyle \frac{\ln x + 1}{\sin x}-\frac{x\ln x \cos x}{\sin^2x}