Bài 13: Ước và bội

\(a,C=5+5^2+5^3+5^4+\cdot\cdot\cdot+5^{20}\)

\(=5\left(1+5+5^2+\cdot\cdot\cdot+5^{19}\right)\)

Ta thấy: \(5\left(1+5+5^2+\cdot\cdot\cdot+5^{19}\right)⋮5\)

nên \(C⋮5\)

\(b,C=5+5^2+5^3+5^4\cdot\cdot\cdot+5^{20}\)

\(=\left(5+5^2\right)+\left(5^3+5^4\right)+\cdot\cdot\cdot+\left(5^{19}+5^{20}\right)\)

\(=5\left(1+5\right)+5^3\left(1+5\right)+\cdot\cdot\cdot+5^{19}\left(1+5\right)\)

\(=5\cdot6+5^3\cdot6+\cdot\cdot\cdot+5^{19}\cdot6\)

\(=6\cdot\left(5+5^3+\cdot\cdot\cdot+5^{19}\right)\)

Ta thấy: \(6\cdot\left(5+5^3+\cdot\cdot\cdot+5^{19}\right)⋮6\)

nên \(C⋮6\)

\(c,C=5+5^2+5^3+5^4+\cdot\cdot\cdot+5^{20}\)

\(=\left(5+5^3\right)+\left(5^2+5^4\right)+\cdot\cdot\cdot+\left(5^{17}+5^{19}\right)+\left(5^{18}+5^{20}\right)\)

\(=5\left(1+5^2\right)+5^2\left(1+5^2\right)+\cdot\cdot\cdot+5^{17}\cdot\left(1+5^2\right)+5^{18}\left(1+5^2\right)\)

\(=5\cdot26+5^2\cdot26+\cdot\cdot\cdot+5^{17}\cdot26+5^{18}\cdot26\)

\(=26\cdot\left(5+5^2+\cdot\cdot\cdot+5^{17}+5^{18}\right)\)

Ta thấy: \(26\cdot\left(5+5^2+\cdot\cdot\cdot+5^{17}+5^{18}\right)⋮13\)

nên \(C⋮13\)

#\(Toru\)

a, ta có
C = 5 + 5^2 + 5^3 + 5^4 + ... + 5^20
=> C = 5 . ( 1 + 5 + 5^2 + 5^3 + ... + 5^19 )
=> C chia hết cho 5
b,
C = 5 + 5^2 + 5^3 + 5^4 + ... + 5^20
=> C = 5 . ( 1 + 5 ) + 5^3 . ( 1 + 5 ) + ... + 5^19 . ( 1 + 5 )
=> C = 5 . 6 + 5^3 . 6 + ... + 5^19 . 6
=> C = 6 . ( 5 + 5^3 + ... + 5^19 )
=> C chia hết cho 6
c,
C = 5 + 5^2 + 5^3 + ... + 5^20
=> C = (5 + 5^2 + 5^3 + 5^4 ) + ... + (5^17 + 5^18 + 5^19 + 5^20 )
=> C = 5 . ( 1 + 5 + 5^2 + 5^3 ) + ... + 5^17 . ( 1+ 5 + 5^2 +5^3)
=> C = 5 . 156 + 5^5 . 156 + ...+ 5^17 . 156
=> C = 5 . 12 . 13 + 5^5 . 12 . 13 + ... + 5^17 . 12 . 13
=> C = 13 . ( 5 . 12 + 5^5 . 12 + ... + 5^17 . 12 )
=> C chia hết cho 13

a) \(20-\left(x+1\right)=10\)

\(\Rightarrow x+1=20-10\)

\(\Rightarrow x+1=10\)

\(\Rightarrow x=10-1\)

\(\Rightarrow x=9\)

b) \(6+10x-4^7:4^5\)

\(\Rightarrow6+10x=4^{7-5}\)

\(\Rightarrow6+10x=4^2\)

\(\Rightarrow6+10x=16\)

\(\Rightarrow10x=16-6\)

\(\Rightarrow10x=10\)

\(\Rightarrow x=\dfrac{10}{10}\)

\(\Rightarrow x=1\)

a: =>x+1=20-10=10

=>x=9

b; =>10x+6=16

=>10x=10

=>x=1

a,  x + 1 = 20 - 10

     x+ 1= 10

     x= 10-1

     x= 9

b,6 +10x = 4²

   ^{6 + 10X = 16}

   ^{10X = 16 - 6}

     ^{10X = 10}

      ^{X= 10 : 10}

     ^{X= 1}

a) \(28.192+72.192+2022\)

\(=192.\left(28+72\right)+2022\)

\(=192.100+2022\)

\(=19200+2022\)

\(=21222\)

_________

d) \(6:\left\{400:\left[500-\left(125+25.7\right)\right]\right\}\)

\(=6:\left\{400:\left[500-300\right]\right\}\)

\(=6:\left\{400:200\right\}\)

\(=6:2\)

\(=3\)

\(#Wendy.Dang\)

\(c,28\cdot192+72\cdot192+2022\)

\(=192\cdot\left(28+72\right)+2022\)

\(=192\cdot100+2022\)

\(=19200+2022\)

\(=21222\)

\(d,6:\left\{400:\left[500-\left(125+25\cdot7\right)\right]\right\}\)

\(=6:\left\{400:\left[500-\left(125+175\right)\right]\right\}\)

\(=6:\left[400:\left(500-300\right)\right]\)

\(=6:\left(400:200\right)\)

\(=6:2\)

\(=3\)

#Toru

a: Ư(8)={1;2;4;8}

Ư(12)={1;2;3;4;6;12}

UC(8;12)={1;2;4}

b: B(16)={0;16;32;...}

B(24)={0;24;48;...}

BC(16,24)={0;48;96;...}

\(c,B\left(12\right)=\left\{0;12;24;36;48;...\right\}\\ B\left(18\right)=\left\{0;18;36;54;72;...\right\}\\ BC\left(12;18\right)=B\left(2^2.3^2\right)=B\left(36\right)=\left\{0;36;72;108;144;....\right\}\\ d,Ư\left(16\right)=\left\{1;2;4;8;16\right\}\\ Ư\left(24\right)=\left\{1;2;3;4;6;8;12;24\right\}\\ ƯC\left(16;24\right)=Ư\left(2^3\right)=Ư\left(8\right)=\left\{1;2;4;8\right\}\\ e,ƯC\left(28;70\right)=Ư\left(2.7\right)=Ư\left(14\right)=\left\{1;2;7;14\right\}\\ BC\left(4;14\right)=B\left(2^2.7\right)=B\left(28\right)=\left\{0;28;56;84;112;140;...\right\}\)

6:

n(n+1)=6

=>n^2+n-6=0

=>(n+3)(n-2)=0

=>n=-3(loại) hoặc n=2(nhận)

4:

Ư(36)={1;-1;2;-2;3;-3;4;-4;6;-6;9;-9;12;-12;18;-18;36;-36}

=>A có 18 phần tử

1:

Ư(100)={1;-1;2;-2;4;-4;5;-5;10;-10;20;-20;25;-25;50;-50;100;-100}

3: 10;50;25

Câu 1: 

\(Ư\left(100\right)=\left\{1;2;4;5;10;25;50;100\right\}\)

Câu 2:

Gọi tập hợp đó là A:
\(A=\left\{0;30;60;90;120;150;...;990\right\}\)

Câu 3: 

Gọi tập hợp đó là B:

\(B=\left\{10;25;50\right\}\)

Đề là gì vậy bạn?

x chia hết cho 8

=>x thuộc {0;8;16;24;...}

mà x<21

nên x thuộc {0;8;16}

Phân tích các số ra thừa số nguyên tố: 

\(4=2^2;7=7;9=3^2\)

=> \(BCNN\left(4;7;9\right)\) = 252

=> Số cần tìm chia hết cho 252 và có chữ số hàng trăm là 7

\(B\left(252\right)\) = \(\left\{0;252;504;756;...\right\}\)

Ta thấy số 756 có chữ số hàng trăm là 7

=> Số cần tìm là 756

18 , 36 , 54 , 72 , 90 💨

80=2^4*5

36=2^2*3^2

140=2^2*5*7

=>ƯCLN(80;36;140)=2^2=4

=>ƯC(80;36;140)={1;-1;2;-2;4;-4}

Ta có bội của 18 là:

\(B\left(18\right)=\left\{0;18;36;54;72;90;...\right\}\)

Ước của 72 là:

\(\text{Ư}\left(72\right)=\left\{1;2;3;4;6;8;9;12;18;36;72\right\}\)

Nên:

\(C=\left\{18;36;72\right\}\)