Question

ajutați. ma repede va rog

Answer 1

g) ( 1 + 2 ori 2² ori 2la a 4-a ) : ( 2 la a 7-a + 1 ) =
     ( 1 + 2 la a 7-a ) : ( 128 + 1 ) =
     ( 1 + 128 ) : 129 =
     129 : 129 =
      1

h) [ ( 3³) totul la a 12-a :  3²   la a 5-a - 56 ] : 5 =
    ( 3 la puterea 36 : 3 la puterea 32 - 56 ) : 5
    ( 3 la puterea a 4-a - 56 ) : 5 =
     ( 81 - 56 ) : 5 = 
     25 : 5 =
     5

i) 8 la a 16-a : ( 2 ori 2 la a 4-a ori 2 la a 10-a ) la a 3-a - 2 la puterea 1 - 2 la puterea 0 =
   2 la puterea 48 : ( 2 la puterea 15 ) totul la a 3-a - 2 - 1 =
   2 la puterea 48 : 2 la puterea 45 - 2 - 1 =
   2 la puterea a 3-a - 2 - 1 =
   8 - 2 - 1 =
   5

Bafta :3

Answer 2

[tex]g) \quad \big(1+2\cdot 2^2\cdot 2^4\big):\big(2^7+1\big) = \big(1+2^{1+2+4}\big):(2^7 +1) = \\ \\ =\big(1+2^7\big):\big(2^7+1\big) = \big(2^7+1\big):\big(2^7+1\big) = 1 \\ \\ \\ h) \quad \Big[(3^3)^{12}:3^{2^5}-56\Big]:5 = \big(3^{3 \cdot 12}}:(3)^{2^5}-56\big):5 = \\ \\ =\big(3^{36}:3^{32}-56\big):5=\big(3^{36-32}-56\big):5 = \big(3^4 - 56\big):5 = \\ \\ =\big(81-56\big):5= 25:5 =5 [/tex]

$i) \quad 8^{16}: \big(2\cdot 2^4\cdot 2^{10}\big)^3-2^1-2^0 = 8^{16} : \big(2^{1+4+10}\big)^3 - 2 - 1 = \\ \\ =2^{3\cdot 16}:\big(2^{15})^3 - 3 = 2^{48} : 2^{15\cdot 3} - 3 = 2^{48}: 2^{45} -3 = 2^{48-45} - 3 = \\ \\ =2^{3}-3 = 8-3 = 5$