Svar 3.4:2
Förberedande kurs i matematik 1
(Skillnad mellan versioner)
(Ny sida: {| width="100%" cellspacing="10px" |a) |width="33%" | <math> \left\{ \eqalign{ x_1&=\sqrt2 \cr x_2&=-\sqrt2 } \right. </math> |b) |width="33%" | <math>x=\ln \left(\displaystyle\frac{\sqrt1...) |
(Ny sida: {| width="100%" cellspacing="10px" |a) |width="33%" | <math> \left\{ \eqalign{ x_1&=\sqrt2 \cr x_2&=-\sqrt2 } \right. </math> |b) |width="33%" | <math>x=\ln \left(\displaystyle\frac{\sqrt1...) |
Nuvarande version
| a) | \displaystyle \left\{ \eqalign{ x_1&=\sqrt2 \cr x_2&=-\sqrt2 } \right. | b) | \displaystyle x=\ln \left(\displaystyle\frac{\sqrt17}{2}-\frac{1}{2}\right) | c) | Saknar lösning |
