Dũng được Tuấn cho số bi là: 21 x \(\dfrac{3}{7}\) = 9 ( viên bi)

Tuấn còn lại số bi là: 21 - 9 = 12 ( viên bi)

Kết luận:.........

A = 2022²⁰²²

A = (2022⁴)⁵⁰⁵.2022² 

A = \(\overline{...6}\)⁵⁰⁵. \(\overline{...4}\)

A = \(\overline{...4}\)

A = 64²⁷ = (64²)¹³.64 = (\(\overline{...6}\))¹³.64 = \(\overline{...6}\) .64 = \(\overline{...4}\) 

B =  19²⁰ = (19²)¹⁰ = \(\overline{...1}\)¹⁰ = \(\overline{...1}\)

C = 114⁴⁴ = (114²)²² = \(\overline{...6}\)¹¹ =  \(\overline{...6}\)

D = 99⁹⁹ = ( 99²)⁴⁹.99 = \(\overline{...1}\)⁴⁹.99 = \(\overline{...9}\)

E = 53⁴⁵ = ( 53⁴)¹¹.53 = \(\overline{...6}\)¹¹. 53  = \(\overline{...6}\).53 = \(\overline{..8}\)

G = 234⁵ = (234²)².234 = \(\overline{..6}\)² .234 = \(\overline{...6}\). 234 = \(\overline{...4}\)

H = (579⁶)³⁵ = (579²)¹⁰⁵ = \(\overline{...1}\)¹⁰⁵ = \(\overline{....1}\)

Lời giải:
$x^5+x-1=(x^5+x^2)-(x^2-x+1)$

$=x^2(x^3+1)-(x^2-x+1)=x^2(x+1)(x^2-x+1)-(x^2-x+1)$

$=(x^2-x+1)[x^2(x+1)-1]=(x^2-x+1)(x^3+x^2-1)$

A = \(-\dfrac{1}{9}\) . (-\(\dfrac{3}{5}\))  - \(\dfrac{5}{6}\) . (-\(\dfrac{3}{5}\)) + \(\dfrac{5}{2}\).(-\(\dfrac{3}{5}\))

A = - \(\dfrac{3}{5}\).( \(-\dfrac{1}{9}\) - \(\dfrac{5}{6}\) + \(\dfrac{5}{2}\))

A = - \(\dfrac{3}{5}\).  \(\dfrac{-2-15+45}{18}\)

A = - \(\dfrac{3}{5}\) . \(\dfrac{28}{18}\)

A = - \(\dfrac{14}{15}\)

   S    = 1 +  3 + 3² + 3³ +...+3⁹

3.S    =       3 + 3² + 3³ +....+3⁹+3¹⁰

3S-S = 3¹⁰ - 1

2S    =  3¹⁰ - 1

 S     = \(\dfrac{3^{10}-1}{2}\)