Ciao a tutti, chi potrebbe darmi una mano?

...e tu in cambio che mi dai ? Un'espressione ? No, grazie  : mi tengo la mano 😉

((2 - 10/11)·((- 4/3)^(-2)/(- 3/4)^3 - (3/4 + 1/2 - 5/3)))^4 - (- 4/5)^(-2) + 4^(-2)=

=(12/11·(3^2/2^4/(- 3^3/2^6) + 5/12))^4 - 25/16 + 1/16=

=(12/11·(- 4/3 + 5/12))^4 - 25/16 + 1/16=

=(12/11·(- 11/12))^4 - 25/16 + 1/16=

=(-1)^4 - 25/16 + 1/16=

=1 - 25/16 + 1/16= - 1/2

(2 - 10/11) = (22/11 - 10/11) = 12/11;

(- 4/3)^-2 = (- 3/4)^2;

4^-2 = (1/4)^2; l'esponente negativo diventa positivo se si inverte la base.

(- 4/5)^-2 = ( - 5/4)^2;

(- 3/4)^2 : (- 3/4)^3 = (- 3/4)^- 1 = (- 4/3)^1, potenze di uguale base:  si sottraggono gli esponenti;

{- 1}^4 = + 1;

{(12/ 11) * [(- 3/4)^2 : (- 3/4)^3 - (3/4 + 1/2 - 5/3)]}^4 - (- 5/4)^2 + (1/4)^2 =

= {(12/ 11) * [(- 3/4)^-1 - (9/12 + 6/12 - 20/12)]}^4 - (+ 25/16) + 1/16 =

= {(12/ 11) * [(- 4/3) - (- 5/12)]}^4 -  25/16 + 1/16 =

= {(12/ 11) * [(- 4/3) + (5/12)]}^4 -  25/16 + 1/16 =

= {(12/ 11) * [(- 16/12 + 5/12)]}^4 - 25/16 + 1/16 =

= {(12/ 11) * [- 11/12]}^4 - 25/16 + 1/16 =

= {- 1}^4 - 25/16 + 1/16 = 

= + 16/16 - 25/16 + 1/16 =

= - 8/16 = - 1/2.

@sara_wgf_uvu  ciao.

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$\small \left\{\left(2-\dfrac{10}{11}\right)·\left[\left(-\dfrac{4}{3}\right)^{-2}÷\left(-\dfrac{3}{4}\right)^3-\left(\dfrac{3}{4}+\dfrac{1}{2}-\dfrac{5}{3}\right)\right]\right\}^4-\left(-\dfrac{4}{5}\right)^{-2}+4^{-2} =$

$\small = \left\{\left(\dfrac{22-10}{11}\right)·\left[\left(-\dfrac{3}{4}\right)^2÷-\dfrac{27}{64}-\left(\dfrac{9+6-20}{12}\right)\right]\right\}^4-\left(-\dfrac{5}{4}\right)^2+\left(\dfrac{1}{4}\right)^2 =$

$\small = \left\{\dfrac{12}{11}·\left[\dfrac{\cancel9^1}{\cancel{16}_1}·-\dfrac{\cancel{64}^4}{\cancel{27}_3}-\left(-\dfrac{5}{12}\right)\right]\right\}^4-\dfrac{25}{16}+\dfrac{1}{16} =$

$\small = \left\{\dfrac{12}{11}·\left[\dfrac{1}{1}·-\dfrac{4}{3}+\dfrac{5}{12}\right]\right\}^4-\dfrac{\cancel{24}^3}{\cancel{16}_2} =$

$\small = \left\{\dfrac{12}{11}·\left[-\dfrac{4}{3}+\dfrac{5}{12}\right]\right\}^4-\dfrac{3}{2} =$

$\small = \left\{\dfrac{12}{11}·\left[\dfrac{-16+5}{12}\right]\right\}^4-\dfrac{3}{2} =$

$\small = \left\{\dfrac{\cancel{12}^1}{\cancel{11}_1}·-\dfrac{\cancel{11}^1}{\cancel{12}_1}\right\}^4-\dfrac{3}{2} =$

$\small = \left\{\dfrac{1}{1}·-\dfrac{1}{1}\right\}^4-\dfrac{3}{2} =$

$\small = \left\{-1\right\}^4-\dfrac{3}{2} =$

$\small = 1-\dfrac{3}{2} =$

$\small = \dfrac{2-3}{2} =$

$\small = -\dfrac{1}{2} $