Ejercicios de Distancia euclídea

Desarrollo:

1. $d({\displaystyle \frac{1}{5}},1)=|1-{\displaystyle \frac{1}{5}}|=|{\displaystyle \frac{5-1}{5}}|={\displaystyle \frac{4}{5}}$

2. $d(-{\displaystyle \frac{1}{3}},{\displaystyle \frac{1}{5}})=|{\displaystyle \frac{1}{5}}-(-{\displaystyle \frac{1}{3}})|=|{\displaystyle \frac{1}{5}}+{\displaystyle \frac{1}{3}}|={\displaystyle \frac{3+5}{15}}={\displaystyle \frac{8}{15}}$

3. $d(-{\displaystyle \frac{1}{5}},-6)=|-6-(-{\displaystyle \frac{1}{5}})|=|-6+{\displaystyle \frac{1}{5}}|=|{\displaystyle \frac{-30+1}{5}}|={\displaystyle \frac{29}{5}}$

4. Si $d(x,-1)<{\displaystyle \frac{1}{3}}$, entonces $|-1-x|<{\displaystyle \frac{1}{3}}$, es decir, $-{\displaystyle \frac{1}{3}}<-1-x<{\displaystyle \frac{1}{3}}$, que sumando $1$ a todos los términos de la desigualdad tenemos que: $$-{\displaystyle \frac{1}{3}}+1<-x<{\displaystyle \frac{1}{3}}+1$$ $${\displaystyle \frac{2}{3}}<-x<{\displaystyle \frac{4}{3}}$$ Multiplicando todos los términos de la igualdad por $-1$ obtenemos que $$-{\displaystyle \frac{2}{3}}>x>-{\displaystyle \frac{4}{3}}$$

Solución:

1. ${\displaystyle \frac{4}{5}}$
2. ${\displaystyle \frac{8}{15}}$
3. ${\displaystyle \frac{29}{5}}$
4. Todos los números reales tales que $-{\displaystyle \frac{2}{3}}>x>-{\displaystyle \frac{4}{3}}$

Ocultar desarrollo y solución