Ho difficoltà con la numero 668

========================================================

$\small \dfrac{\left(\dfrac{144^2}{24^3}\right)^8÷\left[-\left(-\dfrac{2}{3}\right)^3\right]^{-3}}{\left(0,8\overline3\right)^{10}·\left(-0,8\right)^{10}÷\left[-\left(-0,\overline6\right)^4\right]^3} =$

$\small = \dfrac{\left(\dfrac{6^2·24^2}{24^3}\right)^8÷\left[-\left(-\dfrac{8}{27}\right)\right]^{-3}}{\left(\dfrac{83-8}{90}\right)^{10}·\left(-\dfrac{\cancel8^4}{\cancel{10}_5}\right)^{10}÷\left[-\left(-\dfrac{6-0}{9}\right)^4\right]^3} =$

$\small = \dfrac{\left(6^2·24^{2-3}\right)^8÷\left[\dfrac{8}{27}\right]^{-3}}{\left(\dfrac{\cancel{75}^5}{\cancel{90}_6}\right)^{10}·\left(-\dfrac{4}{5}\right)^{10}÷\left[-\left(-\dfrac{\cancel6^2}{\cancel9_3}\right)^4\right]^3} =$

$\small = \dfrac{\left(6^2·24^{-1}\right)^8÷\left[\dfrac{27}{8}\right]^3}{\left(\dfrac{5}{6}\right)^{10}·\left(-\dfrac{4}{5}\right)^{10}÷\left[-\left(-\dfrac{2}{3}\right)^4\right]^3} =$

$\small = \dfrac{\left(\dfrac{6^2}{24}\right)^8·\left[\dfrac{8}{27}\right]^3}{\left(\dfrac{5}{6}·-\dfrac{4}{5}\right)^{10}÷\left[-\dfrac{16}{81}\right]^3} =$

$\small = \dfrac{\dfrac{6^{2·8}}{24^8}·\left[\dfrac{8}{27}\right]^3}{\left(-\dfrac{\cancel{20}^2}{\cancel{30}_3}\right)^{10}·\left[-\dfrac{81}{16}\right]^3} =$

$\small = \dfrac{\dfrac{6^{16}}{4^8·6^8}·\left[\dfrac{8}{27}\right]^3}{\left(-\dfrac{2}{3}\right)^{10}·\left[-\dfrac{81}{16}\right]^3} =$

$\small = \dfrac{\dfrac{6^{16-8}}{4^8}·\left[\dfrac{8}{27}\right]^3}{\left(-\dfrac{2}{3}\right)^{10}·\left[-\dfrac{81}{16}\right]^3} =$

$\small = \dfrac{\dfrac{6^8}{4^8}·\left[\dfrac{8}{27}\right]^3}{\left(-\dfrac{2}{3}\right)^{10}·\left[-\dfrac{81}{16}\right]^3} =$

$\small = \dfrac{\left(\dfrac{\cancel6^3}{\cancel4_2}\right)^8·\left[\dfrac{8}{27}\right]^3}{\left(-\dfrac{2}{3}\right)^{10}·\left[-\dfrac{81}{16}\right]^3} =$

$\small = \dfrac{\left(\dfrac{3}{2}\right)^8·\left[\dfrac{8}{27}\right]^3}{\left(-\dfrac{2}{3}\right)^{10}·\left[-\dfrac{81}{16}\right]^3} =$

$\small = \dfrac{\dfrac{\cancel{6561}^1}{\cancel{256}_1}·\dfrac{\cancel{512}^2}{\cancel{19683}_3}}{\dfrac{\cancel{1024}^1}{\cancel{59049}_1}·-\dfrac{\cancel{531441}^9}{\cancel{4096}_4}} =$

$\small = \dfrac{\dfrac{2}{3}}{-\dfrac{9}{4}} =$

$\small =\dfrac{2}{3}·-\dfrac{4}{9}=$

$\small = -\dfrac{8}{27}$

(144^2/24^3)^8/(- (- 2/3)^3)^(-3)/(((83 - 8)/90)^10·(- 4/5)^10/(- (- 2/3)^4)^3)=

=(2^8·3^4/(2^9·3^3))^8/(2^3/3^3)^(-3)/((5/6)^10·(- 4/5)^10/(- 2^4/3^4)^3)=

=(3/2)^8/(2^3/3^3)^(-3)/((5/6)^10·(- 4/5)^10/(- 2^4/3^4)^3)=

=(3/2)^8/(3^9/2^9)/((5/6)^10·(- 4/5)^10/(- 2^4/3^4)^3)=

=(3/2)^8/(3^9/2^9)/(2^10/3^10/(- 2^4/3^4)^3)=

=2/3/(2^10/3^10/(- 2^4/3^4)^3)=

=2/3/(2^10/3^10/(- 2^12/3^12))=

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

=- 8/27

Prima "brutta frazione" al numeratore...:

[(144^2) /(24^3)]^8 = [(12^2)^2 /(12 * 2)^3]^8 =

= [(12^4) /(12^3 * 2^3)]^8 =

= [(12)^1 /(2^3)]^8 =

= [12/8]^8 = [3/2]^8;

numeratore

[3/2]^8 : [ - (- 2/3)]^(-9) =

[3/2]^8 : [ - (- 3/2)]^(+9) = 

[3/2]^8 : (+ 3/2)^9 =

= [3/2]^(8 - 9) =

= (3/2)^-1 = 2/3;

denominatore:

0,83 (con 3 periodico) =  (83 - 8) / (90) = 75/90 = 5/6;

(5/6)^10....

@ozuna  non ho più tempo ciao.