Câu hỏi của Huỳnh Thị Ngọc Nhung

Question

Chứng tỏ rằng tổng sau chia hết cho 7:

N=2^15-2^12+2^4-2

MIGHFHF · Accepted Answer

$N=2^{15}-2^{12}+2^4-2$

$N=2^{12}.\left(2^3-1\right)+16-2$

$N=2^{12}.7+14$

Vì $\left(2^{12}.7\right)⋮7$

$14⋮7$

$\Rightarrow N⋮7$Câu trả lời: $N=2^{15}-2^{12}+2^4-2$

$N=2^{12}.\left(2^3-1\right)+16-2$

$N=2^{12}.7+14$

Vì $\left(2^{12}.7\right)⋮7$

$14⋮7$

$\Rightarrow N⋮7$

Phương Trâm · Answer

Ta có:

$N=\left(2^{15}-2^{12}\right)+\left(2^4-2\right)$

$N=\left(2^{12}.2^3-2^{12}\right)+\left(2.2^3-2\right)$

$N=2^{12}.\left(2^3-1\right)+2.\left(2^3-1\right)$

$N=\left(2^3-1\right).\left(2^{12}+2\right)$

$N=\left(8-1\right).\left(2^{12}+2\right)$

$N=7.\left(2^{12}+2\right)$ chia hết cho 7 $\left(đpcm\right)$Câu trả lời: Ta có:

$N=\left(2^{15}-2^{12}\right)+\left(2^4-2\right)$

$N=\left(2^{12}.2^3-2^{12}\right)+\left(2.2^3-2\right)$

$N=2^{12}.\left(2^3-1\right)+2.\left(2^3-1\right)$

$N=\left(2^3-1\right).\left(2^{12}+2\right)$

$N=\left(8-1\right).\left(2^{12}+2\right)$

$N=7.\left(2^{12}+2\right)$ chia hết cho 7 $\left(đpcm\right)$

Sáng · Answer

$N=2^{15}-2^{12}+2^4-2$

$\Rightarrow N=\left(2^{15}-2^{12}\right)+\left(2^4-2\right)$

$\Rightarrow N=\left(2^{12}.2^3-2^{12}\right)+\left(2.2^3-2\right)$

$\Rightarrow N=2^{12}.\left(2^3-1\right)+2.\left(2^3-1\right)$

$\Rightarrow N=\left(2^3-1\right).\left(2^{12}+2\right)$

$\Rightarrow N=\left(8-1\right).\left(2^{12}+2\right)$

$\Rightarrow N=7.\left(2^{12}+2\right)$

$\Rightarrow N⋮7$

$\Rightarrowđpcm$Câu trả lời: $N=2^{15}-2^{12}+2^4-2$

$\Rightarrow N=\left(2^{15}-2^{12}\right)+\left(2^4-2\right)$

$\Rightarrow N=\left(2^{12}.2^3-2^{12}\right)+\left(2.2^3-2\right)$

$\Rightarrow N=2^{12}.\left(2^3-1\right)+2.\left(2^3-1\right)$

$\Rightarrow N=\left(2^3-1\right).\left(2^{12}+2\right)$

$\Rightarrow N=\left(8-1\right).\left(2^{12}+2\right)$

$\Rightarrow N=7.\left(2^{12}+2\right)$

$\Rightarrow N⋮7$

$\Rightarrowđpcm$

Đại số lớp 6