Question

Calculați: [(2^6)^4 : 2^3 * (2^6+2^6) - 4^6 : 2^10] : (2^28 - 2^2) + 9
^=la puterea cutare

Answer 1

$[ (2^{6} ) ^{4} : 2^{3} *( 2^{6} + 2^{6})-( 2^{2}) ^{6} : 2^{10} ]$:[tex]( 2^{28}- 2^{2} )+9 [/tex]
$( 2^{24-3} *2* 2^{6} - 2^{12} : 2^{10}) :( 2^{28} - 2^{2} )+9$
$( 2^{21}* 2^{7}- 2^{2} ):( 2^{28} - 2^{2} )+9$
$( 2^{28} - 2^{2} ):( 2^{28} - 2^{2} )+9$
1+9
10

Answer 2

[(2^6)^4:2^3x(2^6+2^6)-4^6:2^10]:(2^28-2^2)+9=(64^4:8x(64+64)-4096:1024):(268435456-4)+9=(16777216:8x128-4):268435452+9=(2097152x128-4): 268435452+9=(268435456-4):268435452+9=268435452:268435452+9=1+9=10