Întrebare
Se dau numerele: a=(0,4×10+0,25×100)÷29+[0,(3)×18+0,1(2)×90]÷17, b=(0,2×5+0,02×100)÷3+(2009 la puterea 2009+2010 la puterea 2010) totul la puterea 0+5×0,8÷4. Sa se calculeze(3a-2b+1) la puterea 2009
Întrebare a fost pusă de: USER2661
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Răspuns (198)
a=(0,4×10+0,25×100)÷29+[0,(3)×18+0,1(2)×90]÷17=
=(4+25)÷29+[1/3×18+11/90×90]÷17=
=29÷29+(6+11)÷17=
=1+1=2
b=(0,2×5+0,02×100)÷3+(2009²⁰⁰⁹+2010²⁰¹⁰)⁰+5×0,8÷4=
=(1+2)÷3+1+4÷4=
=3÷3+1+1=
=1+2=3
(3a-2b+1)²⁰⁰⁹=(3×2-2×3+1)²⁰⁰⁹=1²⁰⁰⁹=1
=(4+25)÷29+[1/3×18+11/90×90]÷17=
=29÷29+(6+11)÷17=
=1+1=2
b=(0,2×5+0,02×100)÷3+(2009²⁰⁰⁹+2010²⁰¹⁰)⁰+5×0,8÷4=
=(1+2)÷3+1+4÷4=
=3÷3+1+1=
=1+2=3
(3a-2b+1)²⁰⁰⁹=(3×2-2×3+1)²⁰⁰⁹=1²⁰⁰⁹=1