Модуль 3(-1.2)+1 целая четыре пятнадцатой+ две третьей

[latex]\bigg|3\cdot (-1,2)+1 \frac{4}{15} \bigg|+ \dfrac{2}{3}=\bigg| -\dfrac{36}{10}+ \dfrac{20}{15} \bigg|+ \dfrac{2}{3} =\bigg| \dfrac{-36\cdot3+20\cdot2}{30} \bigg|+ \dfrac{2}{3} =\\\\\\=\bigg| -\dfrac{68}{30} \bigg|+ \dfrac{2}{3} = \dfrac{68}{30} + \dfrac{20}{30} =\dfrac{88}{30}=2 \frac{28}{30} [/latex]