Question

[nameless]

Tính \(F=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-\frac{3^{16}}{8}\)
Tính \(G=\left(2^3+1\right)\left(2^6+1\right)\left(2^{12}+1\right)-\frac{2^{24}}{7}\)

Answer 1

[Xyz OLM]

a) Ta có F = \(\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-\frac{3^{16}}{8}\)

=> 8F = \(8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-3^{16}\)

=> 8F = \(\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-3^{16}\)

=> 8F = \(\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)-3^{16}\)

=> 8F = \(\left(3^8-1\right)\left(3^8+1\right)-3^{16}=3^{16}-1-3^{16}=-1\)

=> F = -1/8

b) Ta có G = \(\left(2^3+1\right)\left(2^6+1\right)\left(2^{12}+1\right)-\frac{2^{24}}{7}\)

=> 7G = 7(2³ + 1)(2⁶ + 1)(2¹² + 1) - 2²⁴

=> 7G = (2³ - 1)(2³ + 1)(2⁶ + 1)(2¹² + 1) - 2²⁴

=> 7G = (2⁶ - 1)(2⁶ + 1)(2¹² + 1) - 2²⁴

=> 7G = (2¹² - 1)(2¹² + 1) - 2²⁴

=> 7G = 2²⁴ - 1 - 2²⁴

=> 7G = -1

=>  G = -1/7

Answer 2

[Nguyễn Linh Chi]

\(F=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-\frac{3^{16}}{8}\)

<=> \(\left(3^2-1\right)F=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-\left(3^2-1\right)\frac{3^{16}}{8}\)

<=> \(8F=\left(3^4-1\right)\left(3^4+1\right)\left(3^8-1\right)-3^{16}\)

<=> \(8F=\left(3^8+1\right)\left(3^8-1\right)-3^{16}\)

<=> \(8F=\left(3^{16}-1\right)-3^{16}=-1\)

<=> F = -1/8

Câu G làm tương tự