Question

Bài 1: Cho A = 2 + 2^2 + 2^3 + 2^4 +...+ 2^118 + 2^119 + 2^120. Chứng tỏ A chia hết cho 7

Bài 2 :Tính nhanh : 23.78+22.23-15

 

Answer 1

Bài 1 :

\(A=2+2^2+2^3+2^4+...+2^{118}+2^{119}+2^{120}\)

\(\Rightarrow A=2\left(1+2^{ }+2^2\right)+2^4\left(1+2^{ }+2^2\right)+...+2^{118}\left(1+2^{ }+2^2\right)\)

\(\Rightarrow A=2.7+2^4.7+...+2^{118}.7\)

\(\Rightarrow A=7.\left(2+2^4+...+2^{118}\right)⋮7\)

\(\Rightarrow dpcm\)

Bài 2 :

\(...=23\left(78+22\right)-15=23.100-15=2300-15=2285\)

Answer 2

ko bt bài 1

bài 2 là

=23.(78+22)-15

=23.100-15

=2300-15

=2275

hết

Answer 3

Bài 1 :

A = (2 + 2² + 2³) + (2⁴ + 2⁵ + 2⁶) + ... + (2¹¹⁸ + 2¹¹⁹ + 2¹²⁰)

A = 2.(1 + 2 + 4) + 2⁴.(1 + 2 + 4) + ... + 2¹¹⁸.(1 + 2 + 4)

A = 7.(2 + 2⁴ + ...  + 2¹¹⁸)   ⋮  7 (đfcm)

Answer 4

1. ................................ =) A ⋮ 7

2. 23.78 + 22.23 - 15

= 23(78 + 22) - 15

= 23.100 - 15

= 2300 - 15

= 2285

 

Answer 5

Bài 2 :

Đặt B = 23 . (78 + 22) - 15

B = 23 . 100 - 15

B = 2300 - 15

B = 2285