Question

[Hoàng Thành]

Cho A=7^3+7^4+7^5+7^6+...+7^97+7^98 Chứng tỏ A chia hết cho 8

Tìm x

3^47:(189-3×)=3^44

27-3×(5×+2)=6

Lưu ý : dấu ^ là dấu mũ

VD: 3^2 là 3 mũ 2

Answer 1

[Nam Nguyễn]

Bài 1:

\(A=7^3+7^4+7^5+...+7^{97}+7^{98}.\)

\(A=\left(7^3+7^4\right)+\left(7^5+7^6\right)+...+\left(7^{97}+7^{98}\right).\)

\(A=7^3\left(1+7\right)+7^5\left(1+7\right)+...+7^{97}\left(1+7\right).\)

\(A=7^3.8+7^5.8+...+7^{97}.8.\)

\(A=\left(7^3+7^5+...+7^{97}\right).8⋮8\left(đpcm\right).\)

Vậy.....

Bài 2: Tìm x:

\(3^{47}:\left(189-3x\right)=3^{44}.\)

\(189-3x=3^{47}:3^{44}.\)

\(189-3x=27.\)

\(3x=189-27.\)

\(3x=162.\)

\(x=162:3.\)

\(x=54.\)

Vậy.....

\(27-3.\left(5x+2\right)=6.\)

\(3.\left(5x+2\right)=27-6.\)

\(3.\left(5x+2\right)=21.\)

\(5x+2=21:3.\)

\(5x+2=7.\)

\(5x=7-2.\)

\(5x=5.\)

\(x=5:5.\)

\(x=1.\)
Vậy.....

Answer 2

[Ngô Tấn Đạt]

\(A=7^3+7^4+...+7^{98}\\ \Rightarrow A=\left(7^3+7^4\right)+\left(7^5+7^6\right)+....+\left(7^{97}+7^{98}\right)\\ =7^3\left(1+7\right)+7^5\left(1+7\right)+...+7^{97}\left(1+7\right)\\ =7^3.8+7^5.8+...+7^{97}.8\\ =8\left(7^3+7^5+...+7^{97}\right)⋮8\)