Số tổ là Ước của 48 = {1;2;3;4;6;8;12;16;24;48}

Theo đề bài có các cách chia tổ là 4 hoặc 6 tổ

a) \(A=\left\{34;124;128\right\}\)

b) \(B=\left\{315;483\right\}\)

c) \(C=\left\{315\right\}\)

a. \(A=\left\{34;124;128\right\}\)

b. \(B=\left\{315;483;\right\}\)

c. \(C=\left\{315\right\}\)

\(\text{x=20+5=25}\)

a/

\(A=3^2+3^2.2^2+3^2.3^2+3^2.4^2+...+3^2.30^2=\)

\(=3^2\left(1^2+2^2+3^2+...+30^2\right)\)

Đăt biểu thức trong dấu ngoặc là B

\(B=1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+30\left(31-1\right)=\)

\(=1.2+2.3+3.4+30.31-\left(1+2+3+...+30\right)=\)

\(C=1+2+3+...+30=\dfrac{30\left(1+30\right)}{2}=465\)

\(D=1.2+2.3+3.4+...+30.31\)

\(3D=1.2.3+2.3.3+3.4.3+...+30.31.3=\)

\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+30.31.\left(32-29\right)=\)

\(=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-29.30.31+30.31.32=\)

\(=30.31.32\Rightarrow D=\dfrac{30.31.32}{3}=10.31.32\)

\(\Rightarrow A=3^2\left(C-D\right)=3^2\left(10.31.32-465\right)\)

b/

Đặt biểu thức là A

\(\dfrac{A}{2}=\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+\dfrac{9-7}{7.9}+...+\dfrac{39-37}{37.39}=\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{37}-\dfrac{1}{39}=\)

\(=\dfrac{1}{3}-\dfrac{1}{39}=\dfrac{12}{39}\Rightarrow A=\dfrac{2.12}{39}=\dfrac{24}{39}=\dfrac{8}{13}\)

\(A=3.3+6.6+9.9+...+90.90\)

\(A=3^2\left(1+2^2+3^2+...+10^2\right)\)

\(A=9.\dfrac{10.\left(10+1\right)\left(2.10+1\right)}{6}\)

\(A=3.\dfrac{10.11.21}{2}\)

\(A=3465\)

a/

\(A=5\left(1+11+111+...+111...1\right)\) (1999 chữ số 1)

\(A=5\left(\dfrac{10-1}{9}+\dfrac{100-1}{9}+\dfrac{1000-1}{9}+...+\dfrac{1000...0-1}{9}\right)\) (1999 chữ số 0)

\(A=5\left(\dfrac{10+10^2+10^3+...+10^{1999}-1999}{9}\right)\)

Đặt 

\(B=10+10^2+10^3+...+10^{1999}\)

\(10B=10^2+10^3+10^4+...+10^{2000}\)

\(9B=10B-B=10^{2000}-10\)

\(B=\dfrac{10^{2000}-10}{9}=\dfrac{10\left(10^{1999}-1\right)}{9}=\dfrac{10.999...9}{9}=10.111...1\) (1999 chữ số 1)

\(\Rightarrow A=5\left(\dfrac{10.111...1-1999}{9}\right)\) (1999 chữ số 1)

b/

\(C=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{17.19}\)

\(2C=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{19-17}{17.19}=\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{17}-\dfrac{1}{19}=\)

\(=1-\dfrac{1}{19}=\dfrac{18}{19}\Rightarrow C=\dfrac{18}{19}:2=\dfrac{9}{19}\)

Số tự nhiên 2 chữ số là \(\overline{ab}=10a+b\)

Khi thêm chữ số 0 vào giữa : \(\overline{a0b}=100a+b\)

Theo đề ta được :

\(\overline{a0b}=6.\overline{ab}\)

\(\Rightarrow100a+b=6\left(10a+b\right)\)

\(\Rightarrow100a+b=60a+6b\)

\(\Rightarrow40a=5b\)

\(\Rightarrow8a=b\)

\(\Rightarrow\left\{{}\begin{matrix}a=1\\b=8\end{matrix}\right.\)\(\) (vì \(a\in\left\{2;3;4;5;6;7;8;9\right\}\) không phù hợp)

Vậy số đó là 18

132.21= 44.3.21= 44. 63= 63.44

Vậy: 63.44 = 132.21

Ko biết 

\(B\left(9\right)=\left\{0;9;18;27;36;45;...\right\}\)

\(B\left(12\right)=\left\{12;24;36;48;60;72;84;96\right\}\)

\(B\left(4\right)=\left\{0;4;8;12;16;20;24;28;32;36;40;44;48\right\}\)

a) 0;9;18;27;36;45

b) 12;24;36;48;60;72;84;96

c) 0;4;8;12;...;48

Ư(196)={1;2;4;14;49;98;196}

a, Các số là ước 196: 1,4

Ư(100)={1;2;4;5;10;20;25;50;100}

b, các ước có 2 chữ số của 100: 10;20;25;50

c, Ư(24)={1;2;3;4;6;8;12;24}

Ư(48)={1;2;3;4;6;8;12;16;24;48}

Ư(30)={1;2;3;5;6;10;15;30}

a, \(1,4,7\inƯ\left(196\right)\)

b, \(Ư\left(100\right)=\left\{1;2;4;5;10;12;20;50;100\right\};\)

Vậy Ư(100) có 2 chữ số là: \(10;12;50;100\)

c, \(24=2^3.3\)

    \(48=2^4.3\\ 30=2.3.5\)

\(\RightarrowƯCLN\left(24,48,30\right)=2.3=6\\ \RightarrowƯC\left(24;48;30\right)=Ư\left(6\right)=\left\{1;2;3;6\right\}\)