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Motanul14
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Calculati [ 19 supra 6 + ( 1 întreg 1 supra 3) la puterea 9 : (1,(3) la puterea 8 ] x (0,02) la puterea 3
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Răspuns :

Cocirmariadenis
[ 19/6 + ( 4/3)⁹ : (12/9)⁸] x (2/100)³=
=[ 19/6 + 4⁹ / 3⁹ : (4/3)⁸] x ( 1/50)³=
=( 19/6 +4⁹/3⁹ x 3⁸/4⁸) x1/50³=
=( 19/6 +4 /3) x 1/50³=
=( 19/6 +8/6) x 1/50³=
=27/9 x 1/50³=
=0.000036
Didi12342
1 intreg 1/3 = (1×3+1)/3 = 4/3

1,(3) = (13-1)/9 = 12/9 = 4/3

0,02 = 2/100 = 1/50

[19/6 + (4/3)^9 : (4/3)^8] × (1/50)^3 =

= (19/6 + 4/3) × (1/50)^3 =

= (19 + 2×4)/6 × (1/50)^3 =

= (19 + 8)/6 × (1/50)^3 =

= 27/6 × (1/50)^3 =

=9/2 × 125000 =

= 9/250000 =

= (3/500)^2

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