Problems from Absolute value of a real number

Calculate:

  1. $$|-0,75|$$
  2. $$\Big| \dfrac{1}{5}-\dfrac{1}{3}\Big|$$
  3. $$\Big| \dfrac{3\sqrt{8}}{9-\sqrt{6}}\Big|$$
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Development:

  1. $$|-0.75|=max(0.75, -0.75)=0.75$$

  2. $$\Big| \dfrac{1}{5}-\dfrac{1}{3}\Big|=\Big|\dfrac{3-5}{15}\Big|=\Big|\dfrac{-2}{15}\Big|=\dfrac{2}{15}$$

  3. $$\Big| \dfrac{3\sqrt{8}}{9-\sqrt{6}}\Big|=\dfrac{|3\sqrt{8}|}{|9-\sqrt{6}|}=\dfrac{|3|\cdot|\sqrt{8}|}{9-\sqrt{6}}=\dfrac{3\sqrt{8}}{9-\sqrt{6}} $$

Solution:

  1. $$0,75$$
  2. $$\dfrac{2}{15}$$
  3. $$\dfrac{3\sqrt{8}}{9-\sqrt{6}}$$
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Say if the following inequalities are true or false:

  1. $$\Big|\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{1}{5}\Big| < 0,6$$
  2. $$\Big|\dfrac{1}{3}\Big(\dfrac{1}{7}+\dfrac{2}{5}\Big)\Big| < \dfrac{19}{35}$$
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Development:

  1. $$\Big|\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{1}{5}\Big|=\Big|\dfrac{15-20+6}{30}\Big|=\Big|\dfrac{1}{30}\Big|=\dfrac{1}{30}$$

    $$0,6=\dfrac{6}{10}=\dfrac{3}{5}$$, and to be able to compare both fractions, we do $$\dfrac{3}{5}=\dfrac{18}{30}$$, then it is easy to see that $$\dfrac{1}{30} < \dfrac{18}{30} $$

  2. We calculate first: $$$\Big|\dfrac{1}{3}\Big(\dfrac{1}{7}+\dfrac{2}{5}\Big)\Big| = \Big|\dfrac{1}{3}\Big|\cdot \Big|\dfrac{1}{7}+\dfrac{2}{5}\Big|=\dfrac{1}{3}\cdot \Big|\dfrac{5+14}{35}\Big|= \dfrac{1}{3}\cdot\dfrac{19}{35}$$$ and as $$\dfrac{1}{3} < 1$$, we have that $$\dfrac{1}{3}\cdot\dfrac{19}{35} < \dfrac{19}{35}.$$

Solution:

  1. True.
  2. True.
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