\(\dfrac{2^3+2^4+2^5+2^6}{15^2}\)=\(\dfrac{2^3\left(1+2+4+8\right)}{15^2}\)= \(\dfrac{2^3.15}{15^2}\)= \(\dfrac{8}{15}\)

|7x-5y|, |2z-3x| ≥ 0

Để |7x-5y| + |2z-3x| = 0

=> 7x - 5y = 2z - 3x = 0

=> 7x = 5y ; 2z = 3x

=> \(\dfrac{x}{5}\)=\(\dfrac{y}{7}\)

      \(\dfrac{z}{3}\)=\(\dfrac{x}{2}\)

=> \(\dfrac{x}{10}\)=\(\dfrac{y}{14}\)=\(\dfrac{z}{15}\)

=> (\(\dfrac{x}{10}\))² = \(\dfrac{xy}{140}\)= \(\dfrac{yz}{210}\)=\(\dfrac{xz}{150}\)=\(\dfrac{xy+yz+xz}{140+150+210}\)= 4

=> x = 20, y = 28, z = 30

A = 11¹¹ + 111¹¹¹ + 1111¹¹¹¹

A = ...1 + ...1 + ...1

A = ...3

Số chính phương ko tận cùng = 3

=> A ko là số chính phương