Skillnad mellan versioner av "1.6a Lösning 12b"
Från Mathonline
Taifun (Diskussion | bidrag) m |
Taifun (Diskussion | bidrag) m |
||
| Rad 12: | Rad 12: | ||
\end{array}</math> | \end{array}</math> | ||
| − | :<math> \begin{array}{lrcccl} \underline{x \; > \; 0 {\rm :}} & 1 & < & x & < & 5 \\ | + | :<math> \begin{array}{lrcccl} \underline{x \; > \; 0 {\rm :}} & \underline{1 & < & x & < & 5} \\ |
\underline{x \; < \; 0 {\rm :}} & 1 & < & -x & < & 5 \qquad\qquad | \;\; \cdot (-1) \\ | \underline{x \; < \; 0 {\rm :}} & 1 & < & -x & < & 5 \qquad\qquad | \;\; \cdot (-1) \\ | ||
& -1 & > & x & > & -5 \\ | & -1 & > & x & > & -5 \\ | ||
& -5 & < & x & < & -1 | & -5 & < & x & < & -1 | ||
\end{array}</math> | \end{array}</math> | ||
Versionen från 6 september 2014 kl. 13.49
- \[\begin{array}{rcl} \left|\,{5 \over x} + x\,\right| & < & 6 \\ \left|\,{5 + x^2 \over x} \,\right| & < & 6 \\ {|5 + x^2| \over |x|} & < & 6 & | \;\; \cdot |x| \\ |5 + x^2| & < & 6\,|x| \qquad & : \; {\rm VL } \; |5 + x^2| > 0 \\ 5 + x^2 & < & 6\,|x| & | \;\; - 6\,|x| \\ x^2 - 6\,|x| + 5 & < & 0 & | \;\; + 4 \\ x^2 - 6\,|x| + 9 & < & 4 & \\ (\,|x|\,-\,3\,)^2 & < & 4 & | \;\; \sqrt{{\color{White} x}} \\ |\,|x|\,-\,3\,| & < & 2 & | \;\; \sqrt{{\color{White} x}} > 0 \\ -2 \;\; < \;\; |x|\,-\,3\, & < & 2 & | \;\; + 3 \\ 1 \;\; < \;\; |x| & < & 5 \\ \end{array}\]
\[ \begin{array}{lrcccl} \underline{x \; > \; 0 {\rm :}} & \underline{1 & < & x & < & 5} \\ \underline{x \; < \; 0 {\rm :}} & 1 & < & -x & < & 5 \qquad\qquad | \;\; \cdot (-1) \\ & -1 & > & x & > & -5 \\ & -5 & < & x & < & -1 \end{array}\]
