Răspuns:
Explicație pas cu pas:
Se aplica definitia modulului, |a|=a, daca a≥0 si |a|=-a, daca a<0.
[tex]|2-\sqrt{6}|+|3-\sqrt{6} |+|-2\frac{1}{5}|=???\\2=\sqrt{4}<\sqrt{6},~deci~2-\sqrt{6}<0,~deci~ |2-\sqrt{6}|=-(2-\sqrt{6})=-2+\sqrt{6} .\\3=\sqrt{9}>\sqrt{6},~deci~ 3-\sqrt{6}>0,~deci~|3-\sqrt{6}|=3-\sqrt{6}.\\|-2\frac{1}{5}|=-(-2\frac{1}{5})=+2\frac{1}{5}.Atunci,\\|2-\sqrt{6}|+|3-\sqrt{6} |+|-2\frac{1}{5}|=-2+\sqrt{6} +3-\sqrt{6}+2\frac{1}{5}=1+2\frac{1}{5}=3\frac{1}{5}[/tex]
