Question

Va rog mult ajutor cat de urgent!!​

Answer 1

Explicație pas cu pas:

u(15²⁰⁰³) = u(5²⁰⁰³) = u(5) = 5

u(1996²⁰⁰²) = u(6²⁰⁰²) = u(6) = 6

=>

$A = {\bigg[{2}^{ {2}^{3} } \cdot {2}^{u( {15}^{2003} )}\bigg]}^{2} : \bigg[{2}^{u( {1996}^{2002} )} \cdot {( {2}^{6} )}^{3}\bigg] = \\ = {({2}^{8} \cdot {2}^{5} )}^{2} : ({2}^{6} \cdot {2}^{18}) = {({2}^{13})}^{2} : {2}^{6 + 18} =$

$= {2}^{26} : {2}^{24} = {2}^{26 - 24} = {2}^{2} = \bf 4$

u(3²⁰⁰²) = u(3²⁰⁰⁰×3²) = u(u(3⁴)⁵⁰⁰×u(9)) = u(9) = 9

u(3²⁰⁰¹) = u(3²⁰⁰⁰×3) = u(u(3⁴)⁵⁰⁰×u(3)) = u(3) = 3

u(5²⁰⁰³) = u(5) = 5

u(51²⁰⁰³) = u(1²⁰⁰³) = u(1) = 1

u(24²⁰⁰³) = u(4²⁰⁰³) = u(4²⁰⁰²×4) = u((4²)¹⁰⁰¹×u(4)) = u(4) = 4

u(5²⁰⁰³ + 51²⁰⁰³ - 24²⁰⁰³) = u(5 + 1 - 4) = u(2) = 2

=>

$B = \bigg[3^{u(3^{2002})} \cdot 9^{2^{3}} \cdot 3^{u(3^{2001})}\bigg]^{2} : \bigg[27^{2001 : 667} \cdot 3^{u(5^{2003} + 51^{2003} - 24^{2003})} \bigg]^{5} = \\ = \bigg[3^{9} \cdot ( {3}^{2} )^{8} \cdot 3^{3}\bigg]^{2} : \bigg[( {3}^{3} )^{3} \cdot 3^{2} \bigg]^{5} = \bigg(3^{9} \cdot {3}^{16} \cdot 3^{3}\bigg)^{2} : \bigg({3}^{9} \cdot 3^{2} \bigg)^{5} \\ = \Big(3^{9 + 16 + 3}\Big)^{2} : \Big({3}^{9 + 2} \Big)^{5} = \Big(3^{28}\Big)^{2} : \Big({3}^{11} \Big)^{5} = 3^{56} : 3^{55} =$

$= 3^{56 - 55} = \bf 3$

$\implies \bf A > B$