Svar 3.4:3
Förberedande kurs i matematik 1
(Skillnad mellan versioner)
(Ny sida: {| width="100%" cellspacing="10px" |a) |width="50%" | <math>x=-\,\displaystyle\frac{1}{\ln{2}}\pm\sqrt{\left(\displaystyle\frac{1}{\ln{2}}\right)^2-1}</math> |b) |width="50%" | <math>x=\di...) |
(Ny sida: {| width="100%" cellspacing="10px" |a) |width="50%" | <math>x=-\,\displaystyle\frac{1}{\ln{2}}\pm\sqrt{\left(\displaystyle\frac{1}{\ln{2}}\right)^2-1}</math> |b) |width="50%" | <math>x=\di...) |
Nuvarande version
| a) | \displaystyle x=-\,\displaystyle\frac{1}{\ln{2}}\pm\sqrt{\left(\displaystyle\frac{1}{\ln{2}}\right)^2-1} | b) | \displaystyle x=\displaystyle \frac{5}{2} |
| c) | \displaystyle x=1 |
