a; 8 : \(x\) = 2

         \(x\)  = 8 : 2

         \(x\) = 4; \(x\) \(\in\) N*

⇒ \(x\) \(\in\) A = {1; 2; 3; 4}

Tập A có 4 phần tử

b;    \(x\) + 3 < 5 ⇒ \(x\) < 5 - 3 ⇒ \(x< 2\) vì \(x\in\) N 

  ⇒ \(x\) \(\in\) B = {0; 1} Vậy tập B có 2 phần tử

C;  \(x\) - 2 = \(x+2\)

     \(x\) - \(x\) = 2 + 2

        0   =  4 (vô lí)

   C = \(\varnothing\)

Số phần tử của tập C là 0 phần tử

d;    \(x:2=x:4\)

      \(\dfrac{x}{2}\) - \(\dfrac{x}{4}\) = 0

      \(x\) x (\(\dfrac{1}{2}\) - \(\dfrac{1}{4}\)) = 0

        \(x\times\) \(\dfrac{1}{4}\) = 0

        \(x=0\)

D = {0}

Tập D có 1 phần tử

e;    \(x\) + 0 = \(x\) 

        \(x\) = \(x\)

   Vậy E = {0; 1; 2; 3; 4;...}

Tập E có vô số phần tử

\(a)17\cdot85+15\cdot17-120\\ =17\cdot\left(85+15\right)-120\\ =17\cdot100-120\\ =1700-120\\ =1580\\ b)5\cdot7^2-24:2^3\\ =5\cdot49-24:8\\ =245-3\\ =242\\ c)3^3\cdot22-27\cdot19\\ =27\cdot22-27\cdot19\\ =27\cdot\left(22-19\right)\\ =27\cdot3\\ =81\\ d)-\left|-13\right|+\left(-23\right)\\ =-13+\left(-23\right)\\ =-36\\ e)-\left|-13\right|+\left|-25\right|+\left|12\right|\\ =-13+25+12\\ =12+12\\ =24\\ f)23-\left(12-4^2\right)+\left|15\right|\\ =23-\left(12-16\right)+15\\ =23-\left(-4\right)+15\\ =23+4+15\\ =27+15\\ =42\)

\(a)\dfrac{1}{2}x+199=127\cdot36+36\cdot73-30\cdot40\\\dfrac{ 1}{2}x+199=36\cdot\left(127+73\right)-30\cdot40\\ \dfrac{1}{2}x+199=36\cdot200-1200\\ \dfrac{1}{2}x+199=7200-1200\\ \dfrac{1}{2}x+199=6000\\ \dfrac{1}{2}x=6000-199\\ \dfrac{1}{2}x=5801\\ x=5801:\dfrac{1}{2}=11602\)

\(b)\dfrac{3}{2}:x+\dfrac{1}{3}=\dfrac{19}{21}\cdot\dfrac{25}{3}-\dfrac{16}{21}\cdot\dfrac{25}{3}+\dfrac{1}{7}\\ \dfrac{3}{2}:x+\dfrac{1}{3}=\dfrac{25}{3}\cdot\left(\dfrac{19}{3}-\dfrac{16}{3}\right)+\dfrac{1}{7}\\ \dfrac{3}{2}:x+\dfrac{1}{3}=\dfrac{25}{3}+\dfrac{1}{7}\\ \dfrac{3}{2}:x=\dfrac{25}{3}-\dfrac{1}{3}+\dfrac{1}{7}\\ \dfrac{3}{2}:x=8+\dfrac{1}{7}\\ \dfrac{3}{2}:x=\dfrac{57}{7}\\ x=\dfrac{3}{2}:\dfrac{57}{7}\\ x=\dfrac{21}{114}\) 

\(c)\left(x-\dfrac{1}{2025}\right):\dfrac{1}{2}=\dfrac{\dfrac{2023}{3}-\dfrac{2023}{5}-\dfrac{2023}{9}}{\dfrac{2025}{6}-\dfrac{2025}{10}-\dfrac{2025}{18}}\\ \left(x-\dfrac{1}{2025}\right):\dfrac{1}{2}=\dfrac{\dfrac{4046}{6}-\dfrac{4046}{10}-\dfrac{4046}{18}}{2025\left(\dfrac{1}{6}-\dfrac{1}{10}-\dfrac{1}{9}\right)}\\ \left(x-\dfrac{1}{2025}\right):\dfrac{1}{2}=\dfrac{4046\left(\dfrac{1}{6}-\dfrac{1}{10}-\dfrac{1}{9}\right)}{2025\left(\dfrac{1}{6}-\dfrac{1}{10}-\dfrac{1}{9}\right)}\\ \left(x-\dfrac{1}{2025}\right):\dfrac{1}{2}=\dfrac{4046}{2025}\\ x-\dfrac{1}{2025}=\dfrac{4046}{2025}\cdot\dfrac{1}{2}\\ x-\dfrac{1}{2025}=\dfrac{2023}{2025}\\ x=\dfrac{2023}{2025}+\dfrac{1}{2025}\\ x=\dfrac{2024}{2025}\)

\(a)2^3-50:25+13\cdot7=8-2+91\\ =6+91\\ =97\\ b)60-\left[120-\left(42-33\right)\cdot2\right]\\ =60-\left(120-9\cdot2\right)\\ =60-\left(120-18\right)\\ =60-102\\ =-42\\ c)3^{17}:3^{15}+8\cdot3\\ =3^{17-15}+24\\ =3^2+24\\ =9+24\\ =33\\ d)12:\left\{390:\left[500-\left(125+35\cdot7\right)\right]\right\}\\ =12:\left\{390:\left[500-\left(125+245\right)\right]\right\}\\ =12:\left[390:\left(500-370\right)\right]\\ =12:\left(390:130\right)\\ =12:3=4\)

e: \(72^3\cdot49-72^2\cdot9\)

\(=72^2\left(72\cdot49-9\right)\)

\(=5184\cdot3519=18242496\)

f: \(\dfrac{2^3+2^4+2^5}{7^2}=\dfrac{2^3\left(1+2+2^2\right)}{7^2}=\dfrac{8\cdot7}{49}=\dfrac{8}{7}\)

g: \(\dfrac{15^{22}\cdot7^{18}}{7^{20}\cdot15^{21}}=\dfrac{15^{22}}{15^{21}}\cdot\dfrac{7^{18}}{7^{20}}=\dfrac{15}{7^2}=\dfrac{15}{49}\)

7 - n chia hết cho n - 2

=> (-n + 2) + 5 chia hết cho n - 2

=> -(n - 2) + 5 chia hết cho n - 2

=> 5 chia hết cho n - 2

=> n - 2 ∈ Ư(5) = {1; -1; 5; -5}

=> n ∈ {3; 1; 7; -3} 

Bài 3:

\(A=75\left(4^{2004}+4^{2003}+...+4^2+4+1\right)+25\)

Đặt: \(B=4^{2004}+4^{2003}+...+4^2+4+1\)

\(4B=4^{2005}+4^{2004}+...+4^3+4^2+4\\ 4B-B=\left(4^{2005}+4^{2004}+...+4^3+4^2+4\right)-\left(4^{2004}+4^{2003}+...+4^2+4+1\right)\\ 3B=4^{2005}-1\\ B=\dfrac{4^{2005}-1}{3}\)

\(=>A=75\cdot\dfrac{4^{2005}-1}{3}+25\\ =25\left(4^{2005}-1\right)+25\\ =25\cdot\left(4^{2005}-1+1\right)\\ =25\cdot4^{2005}\\ =25\cdot4\cdot4^{2004}\\ =100\cdot4^{2004}\) 

=> A chia hết cho 100 

Bài 1:

a: \(\dfrac{5\cdot3^{11}+4\cdot3^{17}}{3^9\cdot5^2-3^9\cdot2^3}=\dfrac{3^{11}\cdot\left(5+4\cdot3^6\right)}{3^9\left(5^2-2^3\right)}\)

\(=3^2\cdot\dfrac{5+4\cdot729}{25-8}=3^2\cdot\dfrac{2921}{17}=\dfrac{26289}{17}\)

b: \(\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+...+\dfrac{1}{47\cdot50}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{47\cdot50}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{47}-\dfrac{1}{50}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{50}\right)=\dfrac{1}{3}\cdot\dfrac{24}{50}=\dfrac{24}{150}=\dfrac{8}{50}=\dfrac{4}{25}\)

c: \(1+\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{9900}\)

\(=\dfrac{2}{2}+\dfrac{2}{6}+...+\dfrac{2}{9900}\)

\(=2\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{9900}\right)\)

\(=2\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\)

\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=2\left(1-\dfrac{1}{100}\right)=2\cdot\dfrac{99}{100}=\dfrac{99}{50}\)

d: \(\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{99^2}-1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{99}-1\right)\left(\dfrac{1}{2}+1\right)\cdot\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{99}+1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-98}{99}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{100}{99}\)

\(=\dfrac{1}{99}\cdot\dfrac{100}{2}=\dfrac{50}{99}\)

Bài 8:

\(4)-\dfrac{20}{7}:\dfrac{5}{21}\\ =-\dfrac{20}{7}\cdot\dfrac{21}{5}\\ =-4\cdot3\\ =-12\\ 5)-\dfrac{8}{5}:\dfrac{-12}{7}\\ =\dfrac{-8}{5}\cdot\dfrac{-7}{12}\\ =\dfrac{14}{15}\\ 6)\dfrac{-12}{21}:\dfrac{1}{6}\\ =\dfrac{-4}{7}\cdot6\\ =-\dfrac{24}{7}\)

Bài 10:

1: \(4,5\cdot\left(-\dfrac{4}{9}\right)=-\dfrac{9}{2}\cdot\dfrac{4}{9}=-\dfrac{4}{2}=-2\)

2: \(2,4\cdot\left(-3\dfrac{4}{7}\right)=\dfrac{-12}{5}\cdot\dfrac{25}{7}=\dfrac{-12}{7}\cdot\dfrac{25}{5}=-5\cdot\dfrac{12}{7}=-\dfrac{60}{7}\)

3: \(0,2\cdot\dfrac{-15}{4}=\dfrac{1}{5}\cdot\dfrac{-15}{4}=-\dfrac{15}{5}\cdot\dfrac{1}{4}=-\dfrac{3}{4}\)

4: \(\left(-3,5\right):\left(-2\dfrac{4}{5}\right)=\dfrac{3.5}{2\dfrac{4}{5}}=\dfrac{3.5}{2.8}=\dfrac{5}{4}\)

5: \(\dfrac{-5}{23}:\left(-2\right)=\dfrac{5}{23}:2=\dfrac{5}{23\cdot2}=\dfrac{5}{46}\)

6: \(1,25:\left(-3\dfrac{1}{8}\right)=1.25:\dfrac{-25}{8}=1.25\cdot\dfrac{-8}{25}=-\dfrac{10}{25}=-\dfrac{2}{5}\)

Bài 9:

1: \(-3\dfrac{1}{9}\cdot\dfrac{4}{21}=\dfrac{-28}{9}\cdot\dfrac{4}{21}=-\dfrac{28}{21}\cdot\dfrac{4}{9}=-\dfrac{4}{9}\cdot\dfrac{4}{3}=-\dfrac{16}{27}\)

2: \(-\dfrac{3}{4}\cdot2\dfrac{1}{2}=-\dfrac{3}{4}\cdot\dfrac{5}{2}=\dfrac{-15}{8}\)

3: \(-\dfrac{8}{15}\cdot1\dfrac{1}{4}=-\dfrac{8}{15}\cdot\dfrac{5}{4}=\dfrac{-40}{60}=-\dfrac{2}{3}\)

4: \(-\dfrac{11}{15}:1\dfrac{1}{10}=-\dfrac{11}{15}:\dfrac{11}{10}=-\dfrac{11}{15}\cdot\dfrac{10}{11}=-\dfrac{10}{15}=-\dfrac{2}{3}\)

5: \(1\dfrac{1}{5}:\left(-2\dfrac{1}{5}\right)=\dfrac{6}{5}:\dfrac{-11}{5}=\dfrac{6}{5}\cdot\dfrac{5}{-11}=\dfrac{6}{-11}=-\dfrac{6}{11}\)

6: \(\left(-3\dfrac{1}{7}\right):\left(-1\dfrac{6}{49}\right)=\dfrac{-22}{7}:\dfrac{-55}{49}=\dfrac{22}{7}\cdot\dfrac{49}{55}\)

\(=\dfrac{22}{55}\cdot\dfrac{49}{7}=7\cdot\dfrac{2}{5}=\dfrac{14}{5}\)

Bài 8:

1: \(\dfrac{-20}{41}\cdot\dfrac{-4}{5}=\dfrac{20}{5}\cdot\dfrac{4}{41}=4\cdot\dfrac{4}{41}=\dfrac{16}{41}\)

2: \(\dfrac{-24}{5}\cdot\dfrac{15}{-8}=\dfrac{-24}{-8}\cdot\dfrac{15}{5}=3\cdot3=9\)

3: \(\dfrac{-4}{34}\cdot\dfrac{17}{-24}=\dfrac{4}{24}\cdot\dfrac{17}{34}=\dfrac{1}{2}\cdot\dfrac{1}{6}=\dfrac{1}{12}\)

Bài 6:

1: \(2\dfrac{3}{5}-1\dfrac{2}{3}=\dfrac{13}{5}-\dfrac{5}{3}=\dfrac{39}{15}-\dfrac{25}{15}=\dfrac{14}{15}\)

2: \(3\dfrac{3}{7}+2\dfrac{1}{2}=\dfrac{24}{7}+\dfrac{5}{2}=\dfrac{48}{14}+\dfrac{35}{14}=\dfrac{83}{14}\)

3: \(-3\dfrac{1}{2}-2\dfrac{1}{4}=\dfrac{-7}{2}-\dfrac{9}{4}=\dfrac{-14}{4}-\dfrac{9}{4}=-\dfrac{23}{4}\)

4: \(-2\dfrac{1}{2}-3\dfrac{1}{4}=-\dfrac{5}{2}-\dfrac{13}{4}=-\dfrac{10}{4}-\dfrac{13}{4}=-\dfrac{23}{4}\)

5: \(-4\dfrac{1}{2}+2\dfrac{3}{10}=-\dfrac{9}{2}+\dfrac{23}{10}=-\dfrac{45}{10}+\dfrac{23}{10}=-\dfrac{22}{10}=-\dfrac{11}{5}\)

6: \(-6\dfrac{1}{7}-\left(-7\dfrac{1}{6}\right)=-6-\dfrac{1}{7}+7+\dfrac{1}{6}\)

\(=1-\dfrac{1}{7}+\dfrac{1}{6}=\dfrac{6}{7}+\dfrac{1}{6}=\dfrac{43}{42}\)

Bài 7:

\(1)\dfrac{2}{7}+\dfrac{6}{21}-\dfrac{3}{14}\\ =\dfrac{12}{42}+\dfrac{12}{42}-\dfrac{12}{42}\\ =\dfrac{12}{42}\\ =\dfrac{2}{7}\\ 2)-\dfrac{7}{2}+\dfrac{3}{4}-\dfrac{17}{12}\\ =\dfrac{-42}{12}+\dfrac{9}{12}-\dfrac{17}{12}\\ =\dfrac{-50}{12}=\dfrac{-25}{6}\\ 3)\dfrac{1}{12}+\dfrac{1}{4}+\dfrac{2}{3}\\ =\dfrac{1}{12}+\dfrac{3}{12}+\dfrac{8}{12}=\dfrac{12}{12}=1\\ 4)\dfrac{1}{3}+\dfrac{-4}{5}-\dfrac{8}{15}\\ =\dfrac{5}{15}+\dfrac{-12}{15}-\dfrac{8}{15}=-\dfrac{15}{15}=-1\\ 5)\dfrac{2}{3}+\dfrac{-3}{4}+\dfrac{2}{6}\\ =\dfrac{2}{3}+\dfrac{-3}{4}+\dfrac{1}{3}\\ =\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\dfrac{-3}{4}\\ =1+\dfrac{-3}{4}\\ =\dfrac{1}{4}\\ 6\text{ })-\dfrac{5}{18}+\dfrac{5}{45}-\dfrac{9}{6}\\ =-\dfrac{5}{18}+\dfrac{1}{9}-\dfrac{3}{2}\\ =-\dfrac{5}{18}+\dfrac{2}{18}-\dfrac{27}{18}\\ =-\dfrac{30}{18}\\ =-\dfrac{5}{3}\)

Bài 8:

1: \(\dfrac{-20}{41}\cdot\dfrac{-4}{5}=\dfrac{20}{5}\cdot\dfrac{4}{41}=4\cdot\dfrac{4}{41}=\dfrac{16}{41}\)

2: \(\dfrac{-24}{5}\cdot\dfrac{15}{-8}=\dfrac{-24}{-8}\cdot\dfrac{15}{5}=3\cdot3=9\)

3: \(\dfrac{-4}{34}\cdot\dfrac{17}{-24}=\dfrac{4}{24}\cdot\dfrac{17}{34}=\dfrac{1}{2}\cdot\dfrac{1}{6}=\dfrac{1}{12}\)

Bài 6:

1: \(2\dfrac{3}{5}-1\dfrac{2}{3}=\dfrac{13}{5}-\dfrac{5}{3}=\dfrac{39}{15}-\dfrac{25}{15}=\dfrac{14}{15}\)

2: \(3\dfrac{3}{7}+2\dfrac{1}{2}=\dfrac{24}{7}+\dfrac{5}{2}=\dfrac{48}{14}+\dfrac{35}{14}=\dfrac{83}{14}\)

3: \(-3\dfrac{1}{2}-2\dfrac{1}{4}=\dfrac{-7}{2}-\dfrac{9}{4}=\dfrac{-14}{4}-\dfrac{9}{4}=-\dfrac{23}{4}\)

4: \(-2\dfrac{1}{2}-3\dfrac{1}{4}=-\dfrac{5}{2}-\dfrac{13}{4}=-\dfrac{10}{4}-\dfrac{13}{4}=-\dfrac{23}{4}\)

5: \(-4\dfrac{1}{2}+2\dfrac{3}{10}=-\dfrac{9}{2}+\dfrac{23}{10}=-\dfrac{45}{10}+\dfrac{23}{10}=-\dfrac{22}{10}=-\dfrac{11}{5}\)

6: \(-6\dfrac{1}{7}-\left(-7\dfrac{1}{6}\right)=-6-\dfrac{1}{7}+7+\dfrac{1}{6}\)

\(=1-\dfrac{1}{7}+\dfrac{1}{6}=\dfrac{6}{7}+\dfrac{1}{6}=\dfrac{43}{42}\)