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26 tháng 9 2018

Nhận thấy A = 3n + 4n +1 chia hết cho 2 với mọi n tự nhiên, để A chia hết cho 10 ta cần A chia hết cho 5 là đủ.

Nhận xét: 34 \(\equiv\)1 (mod 5), ta sẽ xét các trường hợp: n = 4k, n = 4k+1, n = 4k+2, n = 4k+3 với k là số tự nhiên.

TH1: n = 4k.

A = 34k + 4.(4k) + 1 = 81k + 16k +1 \(\equiv\)1 + k + 1 \(\equiv\)2+k (mod 5)

Để A chia hết cho 5 thì k phải có dạng 5h + 3, với h là số tự nhiên. Vậy n = 4.(5h+3) = 20h +12 thì A chia hết cho 10.

Tương tự với các trường hợp sau bạn giải tiếp nhé!

4 tháng 1 2020

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25 tháng 7 2015

Gọi 2 ps đó là a/b và c/d (ƯCLN (a,b) = 1; ƯCLN (c;d) = 1)

Ta có;

\(\frac{a}{b}+\frac{c}{d}=m\) (m thuộc Z)

=> \(\frac{ad+bc}{bd}=m\)

=> ad + bc = mbd (10

Từ (1) => ad + bc chia hết cho b 

Mà bc chia hết cho b 

=> ad chia hết cho b

Mà (a,b) = 1

=> d chia hết cho b (2)

Từ (1) => ad + bc chia hết cho d 

Mà ad chia hết cho d 

=> bc chia hết cho d

Mà (c,d) = 1

=> b chia hết cho d (3)

Từ (2) và (3) =>bh = d hoặc b = -d (đpcm)