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Rezolvati ex. [(1,1(6)-1/6)] : (-2/3) + 1,5= La puterea 2012 toata paranteza patrata

Răspuns :

[tex]\displaystyle \left[\left(1,1(6)- \frac{1}{6} \right)\right]^{2012}: \left(- \frac{2}{3} \right )+1,5= \\ \\ = \left( \frac{116-11}{90} - \frac{1}{6} \right)^{2012} \cdot \left(- \frac{3}{2} \right)+ \frac{15}{10} = \\ \\ =\left( \frac{105}{90} - \frac{1}{6} \right) ^{2012} \cdot \left(- \frac{3}{2} \right)+ \frac{15}{10} = [/tex]

[tex]\displaystyle = \left( \frac{105}{90} - \frac{15}{90} \right)^{2012} \cdot \left(- \frac{3}{2} \right)+ \frac{15}{10} =\left( \frac{90}{90} \right)^{2012 } \cdot \left(- \frac{3}{2} \right)+ \frac{15}{10} = \\ \\ =1^{2012} \cdot \left(- \frac{3}{2} \right)+ \frac{15}{10} =1 \cdot \left(- \frac{3}{2} \right)+ \frac{15}{10} =- \frac{3}{2} + \frac{15}{10} =- \frac{15}{10} + \frac{15}{10} =0[/tex]