Question

Tìm tập hợp sôa tự nhiên n sao cho:

a) 2n+21 chia hết cho 5

b) 5n-8 chia hết cho 7

c) 7n-20 chia hết cho 13

Answer 1

\(a\)) ( \(2,7,12,17,22,27,...\))

Answer 2

\(b\)) ( \(\text{3,10,24,17,31,39,46 }\))

Answer 3

\(c\)) (\(\text{14,27,40,53,66,79,... }\))

Answer 4

a) Ta có : \(5n-8⋮7\)

\(\Rightarrow5n-8=Ư\left(7\right)=\left\{-7;-1;1;7\right\}\)

\(\Rightarrow5n=\left\{7;9;1;15\right\}\)

\(\Rightarrow n=\left\{3\right\}\)

Answer 5

Mình nhầm câu nãy là câu b

Đây là câu a )

Ta có : \(2n+21⋮5\)

\(\Rightarrow2n+21=Ư\left(5\right)=\left(-5;-1;1;5\right)\)

\(\Rightarrow2n=\left\{-22;-20;-26;-16\right\}\)

\(\Rightarrow n=\left\{-11;-10;-13;-8\right\}\)

Answer 6

c) Ta có : \(7n-20⋮13\)

\(\Rightarrow7n-20=Ư\left(13\right)=\left(-13;-1;1;13\right)\)

\(\Rightarrow7n=\left\{-19;21;7;33\right\}\)

\(\Rightarrow n=\left\{3;1\right\}\)

Answer 7

a)Vì 2n+21 chia hết cho 5

nên 2n+21EB(5)={0;5;10;15;20;25;30;35...}

=>2nE{4;9;14;...}

=>nE{2;7;...}

b)Vì 5n-8 chia hết cho 7

nên 5n-8EB(7)={0;7;14;21;28;35;42;...}

=>5nE{8;15;22;29;36;43;50;...}

=>nE{3;10;...}

c)Vì 7n-20 chia hết cho 13

nên 7n-20EB(13)={0;13;26;39;52;65;120...}

=>7nE{20;33;46;59;72;85;140...}

=>nE{20;...}

Answer 8

a)9,4