Valor absolut d'un nombre real
Calcula:
- $|-0,75|$
- $\Big| \dfrac{1}{5}-\dfrac{1}{3}\Big|$
- $\Big| \dfrac{3\sqrt{8}}{9-\sqrt{6}}\Big|$
$|-0.75|=max(0.75, -0.75)=0.75$
$\Big| \dfrac{1}{5}-\dfrac{1}{3}\Big|=\Big|\dfrac{3-5}{15}\Big|=\Big|\dfrac{-2}{15}\Big|=\dfrac{2}{15}$
$\Big| \dfrac{3\sqrt{8}}{9-\sqrt{6}}\Big|=\dfrac{|3\sqrt{8}|}{|9-\sqrt{6}|}=\dfrac{|3|\cdot|\sqrt{8}|}{9-\sqrt{6}}=\dfrac{3\sqrt{8}}{9-\sqrt{6}} $
- $0,75$
- $\dfrac{2}{15}$
- $\dfrac{3\sqrt{8}}{9-\sqrt{6}}$
Digues si són certes o falses les següents desigualtats:
- $\Big|\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{1}{5}\Big| < 0,6$
- $\Big|\dfrac{1}{3}\Big(\dfrac{1}{7}+\dfrac{2}{5}\Big)\Big| < \dfrac{19}{35}$
- $\Big|\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{1}{5}\Big|=\Big|\dfrac{15-20+6}{30}\Big|=\Big|\dfrac{1}{30}\Big|=\dfrac{1}{30}$
$0,6=\dfrac{6}{10}=\dfrac{3}{5}$, i per poder comparar ambdues fraccions, fem $\dfrac{3}{5}=\dfrac{18}{30}$, de manera que és fàcil veure que $\dfrac{1}{30} < \dfrac{18}{30} $
- Calculem primer: $$\Big|\dfrac{1}{3}\Big(\dfrac{1}{7}+\dfrac{2}{5}\Big)\Big| = \Big|\dfrac{1}{3}\Big|\cdot \Big|\dfrac{1}{7}+\dfrac{2}{5}\Big|=\dfrac{1}{3}\cdot \Big|\dfrac{5+14}{35}\Big|= \dfrac{1}{3}\cdot\dfrac{19}{35}$$ i com $\dfrac{1}{3} < 1$, tenim que $\dfrac{1}{3}\cdot\dfrac{19}{35} < \dfrac{19}{35}.$
- Certa.
- Certa.