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 = 3^4k + 4.(4k) + 1 = 81^k + 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!

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18 lít 

4n+10 chia hết cho 2n+1

 =>

 4n+10=

 ( 2n+1)x2+8

=>(2n+1)x2+8 chia hết cho 2n+1

ma (2n+1)x2 chia hết cho 2

=>8 chia hết cho 2n+1

mà 8 chia hết cho:1;2;4;8

=>vay n=0

4n + 10 chia hết cho 2n+1 thì 2(4n +10) cũng chia hết cho 4(2n+1)

xét hiệu ta có 8n+20 - 8n-4 = 16

vì 4n+10 chia hết cho 2n+1 nên 2(4n+10) chia hết cho 2n+1

4(2n+1) chia hết cho 2n+1

=> 16 chia hết cho 2n+1

vậy 2n+1 thuộc ước của 16

a)4n-7chia hết cho n-1<=>4(n-1)-3chia hết cho n-1=>3chia hết cho n-1 mà n thuộc N

=>n-1 thuộc{-1;1;3}

=>n thuộc {1;2;4}

b)10-2n chia hết cho n-2<=>14-2(n-2) chia hết cho n-2 =>14 chia hết cho n-2 mà n thuộc N

=>n-2 thuộc {-2;-1;1;2;7;14}

=>n thuộc {0;1;3;4;9;16}

c)5n-8 chia hết cho 4-n <=>5(4-n)-28 chia hết cho n-4=>28chia hết cho n-4 mà n thuộc N

=>n-4 thuộc {-4;-2;-1;1;2;4;7;14;28}

=>n thuộc{0;2;3;5;6;8;11;18;32}

d)n²+3n+6 chia hết cho n-3<=>-n(n-3)+6 chia hết cho n-3=>6 chia hết cho n-3 mà n thuộc N

=>n-3 thuộc{-3;-2;-1;1;2;3;6}

=>n thuộc{0;1;2;4;5;6;9}

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Đúng ko bạn hiền 

Answer:

\(B=\frac{10n-3}{4n-10}\)

\(=\frac{5.\left(2n-5\right)+22}{2.\left(n-5\right)}\)

\(=\frac{5}{2}+\frac{22}{2.\left(2n-5\right)}\)

\(=\frac{5}{2}+\frac{11}{2n-5}\)

Mà để B đạt giá trị lớn nhất thì \(\frac{11}{2n-5}\) đạt giá trị lớn nhất

Mà ta có: 11 > 0 thì \(\frac{11}{2n-5}\) đạt giá trị lớn nhất khi: 

2n - 5 > 0 và đạt giá trị nhỏ nhất khi: \(2n-5=1\Rightarrow2n=6\Rightarrow n=3\)

Tương tự: Giá trị lớn nhất là: \(11+\frac{5}{2}=13,5\)

Vậy giá trị lớn nhất của biểu thức \(B=13,5\) khi \(n=3\)