Question

Va rog nu va bateti joc, dau coroana. Am nevoie urgent! ❤️​

Answer 1

Explicație pas cu pas:

$A = \dfrac{12^{(12} }{48} + \dfrac{102^{(102} }{408} + \dfrac{1002^{(1002} }{4008} + \dfrac{10002^{(10002} }{40008} =$

$= \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{4}{4} = \bf 1$

$B = 1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + ... + \dfrac{1}{2008} + \dfrac{1}{2} + \dfrac{2}{3} + \dfrac{3}{4} + ... + \dfrac{2007}{2008} =$

$= 1 + \Big(\dfrac{1}{2} + \dfrac{1}{2} \Big) + \Big(\dfrac{1}{3} + \dfrac{2}{3} \Big) + \Big(\dfrac{1}{4} + \dfrac{3}{4} \Big) + ... + \Big(\dfrac{1}{2008} + \dfrac{2007}{2008} \Big)$

$= 1 + \underbrace{\dfrac{2}{2} + \dfrac{3}{3} + \dfrac{4}{4} + ... + \dfrac{2008}{2008}}_{2007 \ \ termeni} = 1 + \underbrace{1 + 1 + 1 + ... + 1}_{2007 \ \ termeni}$

$= 1 + 2007 = \bf 2008$

=>

$B - A = 2008 - 1 = \bf 2007$

${A}^{B} = {1}^{2008} = \bf 1$

${(B - 2002)}^{A - 1} = {(2008 - 2002)}^{1 - 1} = {6}^{0} = \bf 1$