Question

Dau coroana va rog io din toata inima mea sa rezolve cnv ex asta

Answer 1

Răspuns:

3

Explicație pas cu pas:

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

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

u(5²⁰⁰³+51²⁰⁰³-24²⁰⁰³) = u[(5²⁰⁰³)+u(51²⁰⁰³)-u(24²⁰⁰³)] = u[(5²⁰⁰³)+u(1²⁰⁰³)-u(4²⁰⁰³)] = u[(5)+u(1)-u((4²)¹⁰⁰¹×4)] = u[5+1-u(4)] = u(6-4) = u(2) = 2

=>

$B = {\Big( {3}^{9} \cdot {9}^{ {2}^{3}} \cdot {3}^{3}\Big)}^{2} : {\Big({27}^{2001 : 667} \cdot {3}^{2}\Big)}^{5} =$

$= {\Big[ {3}^{9} \cdot {( {3}^{2} )}^{8} \cdot {3}^{3}\Big ]}^{2} : {\Big[ {( {3}^{3} )}^{3} \cdot {3}^{2}\Big]}^{5}$

$= {\Big( {3}^{9} \cdot {3}^{16} \cdot {3}^{3}\Big )}^{2} : {\Big({3}^{9} \cdot {3}^{2}\Big)}^{5}$

$= {\Big( {3}^{9 + 16 + 3}\Big )}^{2} : {\Big({3}^{9 + 2}\Big)}^{5} = {\Big( {3}^{28}\Big )}^{2} : {\Big({3}^{11}\Big)}^{5}$

$= {3}^{28 \cdot 2} : {3}^{11 \cdot 5} = {3}^{56} : {3}^{55} = {3}^{56 - 55} = \bf 3$