Question

Efectuati :
(2^ 38:2 ^10 +6^ 23:613):[2 ^10* 3^10+(2^14)^2]
3^40:[3^20*3^18+(3^5*3^14)^@+(4^20:4^-1)^38]
Ps:^=putere

Answer 1

$(2^{28} + 6^{10} ):( 2^{10} * 3^{10} + 2^{28} )= ( 2^{28} + 6^{10} ):( 6^{10} + 2^{28} )=1$

2. [tex] 3^{40} :[ 3^{20} * 3^{18} +( 3^{19} )^{2} +( 4^{21} ) ^{38} ] = 3^{40} :[ 3^{38} + 3^{38} + (2^{2} ) ^{21} ] ^{38} } [/tex]$3^{40}:[ 2*3^{38} + (2^{42} ) ^{38} ]=$