Tìm số tự nhiên n để \(3^n+4n+1⋮10\)
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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!
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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
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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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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 \(n=3\)