Espressioni con le proprietà delle potenze.

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

$\small \left\{\left[\left(4^0+4\right)·5^3\right]^2 ÷ 5^6-2·\left(3^4+5^0+3·2\right) ÷ 29\right\}·\left(5+5^0\right)+2^3 ÷ 2 =$

$\small =\left\{\left[\left(1+4\right)·5^3\right]^2 ÷ 5^6-2·\left(81+1+6\right) ÷ 29\right\}·\left(5+1\right)+2^{3-1} =$

$\small =\left\{\left[5·5^3\right]^2 ÷ 5^6-2·88 ÷ 29\right\}·6+2^2 =$

$\small =\left\{\left[5^{1+3}\right]^2 ÷ 5^6- \dfrac{176}{29}\right\}·6+4 =$

$\small =\left\{\left[5^4\right]^2 ÷ 5^6- \dfrac{176}{29}\right\}·6+4 =$

$\small =\left\{5^{4·2} ÷ 5^6- \dfrac{176}{29}\right\}·6+4 =$

$\small =\left\{5^8 ÷ 5^6- \dfrac{176}{29}\right\}·6+4 =$

$\small =\left\{5^{8-6}- \dfrac{176}{29}\right\}·6+4 =$

$\small =\left\{5^2- \dfrac{176}{29}\right\}·6+4 =$

$\small =\left\{25- \dfrac{176}{29}\right\}·6+4 =$

$\small =\dfrac{725-176}{29}·6+4 =$

$\small =\dfrac{549}{29}·6+4 =$

$\small =\dfrac{3294}{29}+4 =$

$\small =\dfrac{3294+116}{29} =$

$\small =\dfrac{3410}{29} $

(((4^0 + 4)·5^3)^2/5^6 - 2·(3^4 + 5^0 + 3·2)/29)·(5 + 5^0) + 2^3/2=

=(((1 + 4)·5^3)^2/5^6 - 2·(3^4 + 1 + 6)/29)·(5 + 1) + 4=

=(5^8/5^6 - 2·(81 + 1 + 6)/29)·6 + 4=

=(25 - 176/29)·6 + 4=

=549/29·6 + 4=

=3410/29

Secondo me hai sbagliato a digitare qualcosa...

(((1+ 4)*5^3))^2 / 5^6 - 2*(3^4 + 1 + 3*2)/29)) * (5 + 1) + 2^3/2

((5^(8-6) - 2*(87/29)) * 6 + 4

(25 - 2*3) * 6 + 4 

19*6 + 4 

118

50+50+8*9 : ((56+3(5+12-17)-20)

50 + 8*9/36

52

110^3 :(35*(7-5))^3

110^3/70^3

(11/7)^3 

3,8805...

9^20 : 9^(8+2+3) : (3^13+49*0)

9^(20-15) : 3^13

9^5 : 3^13

(3^2)^5 : 3^13

3^(2*5) : 3^13

3^(10-13)

3^-3

(3^2+1)^2 : 10 +((81-74) : 7^2 +2)^2

100 : 10 + (7/7^2+2)^2

10+(15/7)^2

la quantità 

vale zero ed essendo il denominatore di un quoziente da luogo ad una quantità indeterminata 

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

$\small \left\{(5-1·3)^2+2^6÷(56-6·9)^2-\left(2^2·3^2\right)÷(4+8·5-5·7)\right\}÷2^2=$

$\small =\left\{(5-3)^2+2^6÷(56-54)^2-(2·3)^2÷(4+40-35)\right\}÷4=$

$\small =\left\{2^2+2^6÷2^2-6^2÷9\right\}÷4=$

$\small =\left\{4+2^{6-2}-36÷9\right\}÷4=$

$\small =\left\{4+2^4-4\right\}÷4=$

$\small =\left\{4+16-4\right\}÷4=$

$\small =16÷4=$

$\small =4$

{ [ (4⁰+4) •5³ ]² :5⁶-2• (3⁴ +5⁰ +3 •2) : 29 }•   (5+5⁰) +2³ : 2=

= {[(1 + 4) * 5^3]^2 : 5^6 - 2 * (81 + 1 + 6): 29} * (5 + 1) + 2^2 =

= {5 * 5^3]^2 : 5^6 - 2 * 88 : 29}* 6 + 4 =

= {[5^4]^2 : 5^6 - 176 : 29} * 6 + 4 =

= {5^8 : 5^6 - 176 : 29} * 6 + 4 =;

[il numero 29 sembra sbagliato, non è un divisore di  176. Come facciamo?

Andrebbe bene 87 : 29,  bisogna eliminare 5^0 = 1 dentro la tonda;

con la correzione, avremo ....

2 * (3^4 + 3 * 2) : 29 = 2 * (81 + 6) : 29 =

= 2 * 87 : 29 = 174 : 29 = 6  ]. 

Al posto di 176 mettiamo 174:

= {5^2 - 174 : 29} * 6 + 4 = 

= {25 - 6} * 6 + 4 =

= 19 * 6 + 4 = 114 + 4 = 118.

Ciao @lya

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

$\small (25·2+10·5)+9·8÷\{56+3·[5+6·(3·4-10)-17]-5·4\}=$

$\small =50+50+72÷\{56+3·[5+6·(12-10)-17]-20\}=$

$\small =100+72÷\{56+3·[5+6·2-17]-20\}=$

$\small =100+72÷\{56+3·[5+12-17]-20\}=$

$\small =100+72÷\{56+3·0-20\}=$

$\small =100+72÷\{56+0-20\}=$

$\small =100+72÷36=$

$\small =100+2=$

$\small =102$