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7:(5/2-2/5)+11/30:1 întreg 2/3+(-1)la puterea 2005
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Răspuns :

Renatemambouko
7:(5/2-2/5)+11/30:1 întreg 2/3+(-1)²⁰⁰⁵=
=7:(25-4)/10+11/30:5/3-1=
=7×10/21+11/30×3/5-1=
=10/3+11/50-1=
=500/150+33/150-150/150=
=383/150


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