Espressioni con potenze n.280

9/2 : [(7/8 + 2/8)^3 : (9/8)^2 ]^2 + (1/3) * (1/9) * (1/27) : (1/81) =

= 9/2 : [(9/8)^3 : (9/8)^2 ]^2 + (1/3) * (1/9) * (1/27) * (81/1) =

 Spiegazioni:  {(9/8)^3 : (9/8)^2 = (9/8)^(3 - 2) ; potenze di uguale base, si sottraggono gli esponenti;}

{moltiplicazioni:    81 al numeratore si semplifica con 3 * 27 = 81 al denominatore; resta 1/9;

(1/3) * (1/9) * (1/27) * (81/1) = 1/9}.

Riprendendo l'espressione:

= 9/2 : [(9/8)^3 : (9/8)^2 ]^2 + (1/3) * (1/9) * (1/27) * (81/1) =

= 9/2 : [(9/8)^1]^2 + 1/9 =

= 9/2 : (9/8)^2 + 1/9 =

= 9/2 : 81/64 + 1/9 =

= 9/2 * 64/81 + 1/9 =

= 32/9 + 1/9 =

= 33/9 =

= 11/3.

@tatia  ciao.

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$\small \dfrac{9}{2} : \left[\left(\dfrac{7}{8}+\dfrac{1}{4}\right)^3 : \left(\dfrac{9}{8}\right)^2\right]^2+\dfrac{1}{3}·\dfrac{1}{9}·\dfrac{1}{27} : \left(\dfrac{1}{9}\right)^2=$

$\small =\dfrac{9}{2} : \left[\left(\dfrac{7+2}{8}\right)^3 : \left(\dfrac{9}{8}\right)^2\right]^2+\dfrac{1}{729} : \dfrac{1}{81}=$

$\small =\dfrac{9}{2} : \left[\left(\dfrac{9}{8}\right)^3 : \left(\dfrac{9}{8}\right)^2\right]^2+\dfrac{1}{729}·\dfrac{81}{1}=$

$\small =\dfrac{9}{2} : \left[\left(\dfrac{9}{8}\right)^{3-2}\right]^2+\dfrac{1}{\cancel{729}_9}·\dfrac{\cancel{81}^1}{1}=$

$\small =\dfrac{9}{2} : \left[\left(\dfrac{9}{8}\right)^1\right]^2+\dfrac{1}{9}·\dfrac{1}{1}=$

$\small =\dfrac{9}{2} : \left(\dfrac{9}{8}\right)^{1·2}+\dfrac{1}{9}=$

$\small =\dfrac{9}{2} : \left(\dfrac{9}{8}\right)^2+\dfrac{1}{9}=$

$\small =\dfrac{9}{2} : \dfrac{81}{64}+\dfrac{1}{9}=$

$\small =\dfrac{9}{2}·\dfrac{64}{81}+\dfrac{1}{9}=$

$\small =\dfrac{\cancel9^1}{\cancel2_1}·\dfrac{\cancel{64}^{32}}{\cancel{81}_9}+\dfrac{1}{9}=$

$\small =\dfrac{1}{1}·\dfrac{32}{9}+\dfrac{1}{9}=$

$\small =\dfrac{32}{9}+\dfrac{1}{9}=$

$\small =\dfrac{\cancel{33}^{11}}{\cancel9_3}=$

$\small =\dfrac{11}{3}$