\(b.\left(1+\frac12\right)\left(1+\frac13\right)\left(1+\frac14\right)\ldots\left(1+\frac{1}{2023}\right)\)

\(=\frac32\cdot\frac43\cdot\frac54\cdot\ldots\cdot\frac{2024}{2023}\)

\(=\frac{3\cdot4\cdot5\cdot\ldots\cdot2024}{2\cdot3\cdot4\cdot\ldots\cdot2023}\)

\(=\frac{2024}{2}=1012\)

\(c.D=\frac{5}{6\cdot37}+\frac{1}{6\cdot43}+\frac{6}{7\cdot43}+\frac{10}{7\cdot59}\)

\(D=7\cdot\left(\frac{5}{37\cdot42}+\frac{1}{42\cdot43}+\frac{6}{43\cdot49}+\frac{10}{49\cdot59}\right)\)

\(D=7\cdot\left(\frac{1}{37}-\frac{1}{42}+\frac{1}{42}-\frac{1}{43}+\frac{1}{43}-\frac{1}{49}+\frac{1}{49}-\frac{1}{59}\right)\)

\(D=7\cdot\left(\frac{1}{37}-\frac{1}{59}\right)\)

\(D=7\cdot\frac{22}{2183}\)

\(D=\frac{154}{2183}\)

\(A=1+3+3^2+...+3^{2016}\)

\(3A=3.\left(1+3+3^2+...+3^{2016}\right)\)

\(3A=3+3^2+3^3+...+3^{2017}\)

\(3A-A=\left(3+3^2+3^3+...+3^{2017}\right)-\left(1+3+3^2+...+3^{2016}\right)\)

\(2A=3^{2017}-1\)

\(A=\left(3^{2017}-1\right):2\)

\(B=1+6+6^2+...+6^{200}\)

\(6B=6.\left(1+6+6^2+...+6^{200}\right)\)

\(6B=6+6^2+6^3+...+6^{201}\)

\(6B-B=\left(6+6^2+6^3+...+3^{201}\right)-\left(1+6+6^2+...+6^{200}\right)\)

\(5B=6^{201}-1\)

\(B=\left(6^{201}-1\right):5\)

\(3^{x-2}.4=324\)

\(3^{x-2}=324:4\)

\(3^{x-2}=81\)

\(3^{x-2}=3^4\)

\(x-2=4\)

\(x=4+2\)

\(x=6\)

\(2x< 20\)

\(\Rightarrow x=\left\{0;1;2;3;4;5;6;7;8;9\right\}\)

a)\(11\frac{1}{4}-\left(2\frac{5}{7}+5\frac{1}{4}\right)\)

\(=\frac{45}{4}-\left(\frac{19}{7}+\frac{21}{4}\right)\)

\(=\frac{45}{4}-\left(\frac{76}{28}+\frac{147}{28}\right)\)

\(=\frac{45}{4}-\frac{223}{28}\)

\(=\frac{315}{28}-\frac{223}{28}\)

\(=\frac{23}{7}\)

b) \(\left(8\frac{5}{11}+3\frac{5}{8}\right)-3\frac{5}{11}\)

   \(=\left(\frac{93}{11}+\frac{29}{8}\right)-\frac{38}{11}\)

   \(=\left(\frac{744}{88}+\frac{319}{88}\right)-\frac{38}{11}\)

   \(=\frac{1063}{88}-\frac{38}{11}=\frac{1063}{88}-\frac{304}{88}\)

    \(=\frac{69}{8}\)

1a) \(11\frac{1}{4}-\left(2\frac{5}{7}+5\frac{1}{4}\right)\)

\(=11\frac{1}{4}-2\frac{5}{7}-5\frac{1}{4}\)

\(=11\frac{1}{4}-5\frac{1}{4}-2\frac{5}{7}\)

\(=6-2\frac{5}{7}=3\frac{2}{7}=\frac{23}{7}\)

2a) \(x\div8,25=\left(-2,2\right)\)

\(\Leftrightarrow x=-18,15\)

\(a,\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\)

=> \(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\)

=> \(x=\frac{3}{10}:\frac{2}{3}=\frac{9}{20}\)

Vậy \(x\in\left\{\frac{9}{20}\right\}\)

\(b,x+\frac{1}{4}=\frac{4}{3}\)

=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)

Vậy \(x\in\left\{\frac{13}{12}\right\}\)

\(c,\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)

=> \(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}=\frac{5}{14}\)

=> \(x=\frac{5}{14}:\frac{3}{5}=\frac{25}{42}\)

Vậy \(x\in\left\{\frac{25}{42}\right\}\)

\(d,\left|x+5\right|-6=9\)

=> \(\left|x+5\right|=9+6=15\)

=> \(\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=15-5=10\\x=-15-5=-20\end{matrix}\right.\)

Vậy \(x\in\left\{10;-20\right\}\)

\(e,\left|x-\frac{4}{5}\right|=\frac{3}{4}\)

=> \(\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{4}\\x-\frac{4}{5}=-\frac{3}{4}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{3}{4}+\frac{4}{5}=\frac{31}{20}\\x=-\frac{3}{4}+\frac{4}{5}=\frac{1}{20}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{31}{20};\frac{1}{20}\right\}\)

\(f,\frac{1}{2}-\left|x\right|=\frac{1}{3}\)

=> \(\left|x\right|=\frac{1}{2}-\frac{1}{3}\)

=> \(\left|x\right|=\frac{1}{6}\)

=> \(\left[{}\begin{matrix}x=\frac{1}{6}\\x=-\frac{1}{6}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{1}{6};-\frac{1}{6}\right\}\)

\(g,x^2=16\)

=> \(\left|x\right|=\sqrt{16}=4\)

=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

vậy \(x\in\left\{4;-4\right\}\)

\(h,\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)

=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)

=> \(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)

Vậy \(x\in\left\{\frac{5}{6}\right\}\)

\(i,3^3.x=3^6\)

\(x=3^6:3^3=3^3=27\)

Vậy \(x\in\left\{27\right\}\)

\(J,\frac{1,35}{0,2}=\frac{1,25}{x}\)

=> \(x=\frac{1,25.0,2}{1,35}=\frac{5}{27}\)

Vậy \(x\in\left\{\frac{5}{27}\right\}\)

\(k,1\frac{2}{3}:x=6:0,3\)

=> \(\frac{5}{3}:x=20\)

=> \(x=\frac{5}{3}:20=\frac{1}{12}\)

Vậy \(x\in\left\{\frac{1}{12}\right\}\)

Câu a:

1/3x = - 5/6

x = -5/6 : 1/3

x = -5/6 x 3

x = - 5/2

Vậy x = - 5/2

Câu b:

x : - 3/2 = - 4/5

x = - (4/5) x (-3/2)

x = 6/5

Vậy x = 6/5

Câu c:

x + 5/2 = -1 1/4

x + 5/2 = - 5/4

x = - 5/4 - 5/2

x = - 5/4 - 10/4

x = - 15/4

Vậy x = - 15/4

Câu d:

6/7 - x = - 1/14

x = 6/7 + 1/14

x = 12/14 + 1/14

x = 13/14

Vậy x = 13/14

Bài 1:

a) \(\left(\dfrac{3}{8}+\dfrac{-3}{4}+\dfrac{7}{12}\right):\dfrac{5}{6}+\dfrac{1}{2}\)

\(=\left(\dfrac{9}{24}+\dfrac{-18}{24}+\dfrac{14}{24}\right):\dfrac{5}{6}+\dfrac{1}{2}\)

\(=\dfrac{5}{24}:\dfrac{5}{6}+\dfrac{1}{2}\)

\(=\dfrac{5}{24}.\dfrac{6}{5}+\dfrac{1}{2}\)

\(=\dfrac{1}{4}+\dfrac{1}{2}\)

\(=\dfrac{1}{4}+\dfrac{2}{4}\)

\(=\dfrac{3}{4}\)

b) \(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\)

\(=\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\)

\(=\left(\dfrac{1}{2}+\dfrac{4}{5}\right)+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)\)

\(=\dfrac{1}{2}+\dfrac{4}{5}\)

\(=\dfrac{5}{10}+\dfrac{8}{10}\)

\(=\dfrac{9}{5}\)

c) \(6\dfrac{5}{12}:2\dfrac{3}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)

\(=\dfrac{77}{12}:\dfrac{11}{4}+\dfrac{42}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)

\(=\dfrac{77}{12}.\dfrac{4}{11}+\dfrac{42}{4}.\left(\dfrac{5}{15}+\dfrac{3}{15}\right)\)

\(=\dfrac{7}{3}+\dfrac{42}{4}.\dfrac{8}{15}\)

\(=\dfrac{7}{3}+\dfrac{14.2}{1.3}\)

\(=\dfrac{7}{3}+\dfrac{28}{3}\)

\(=\dfrac{35}{3}\)

d) \(\left(\dfrac{7}{8}-\dfrac{3}{4}\right).1\dfrac{1}{3}-\dfrac{2}{7}.\left(3,5\right)^2\)

\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right).\dfrac{4}{3}-\dfrac{2}{7}.12\dfrac{1}{4}\)

\(=\dfrac{1}{8}.\dfrac{4}{3}-\dfrac{2}{7}.\dfrac{49}{4}\)

\(=\dfrac{1}{6}-\dfrac{7}{2}\)

\(=\dfrac{1}{6}-\dfrac{21}{6}\)

\(=\dfrac{-10}{3}\)

e) \(\left(\dfrac{3}{5}+0,415-\dfrac{3}{200}\right).2\dfrac{2}{3}.0,25\)

\(=\left(\dfrac{3}{5}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)

\(=\left(\dfrac{120}{200}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)

\(=1.\dfrac{8}{3}.\dfrac{1}{4}\)

\(=\dfrac{2}{3}\)

f) \(\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)

\(=\dfrac{5}{16}:\dfrac{1}{8}-\left(\dfrac{9}{4}-\dfrac{3}{5}\right).\dfrac{10}{11}\)

\(=\dfrac{5}{16}.\dfrac{8}{1}-\left(\dfrac{45}{20}-\dfrac{12}{20}\right).\dfrac{10}{11}\)

\(=\dfrac{5}{2}-\dfrac{33}{20}.\dfrac{10}{11}\)

\(=\dfrac{5}{2}-\dfrac{3}{2}\)

\(=\dfrac{2}{2}=1\)

g) \(0,25:\left(10,3-9,8\right)-\dfrac{3}{4}\)

\(=\dfrac{1}{4}:\dfrac{1}{2}-\dfrac{3}{4}\)

\(=\dfrac{1}{4}.\dfrac{2}{1}-\dfrac{3}{4}\)

\(=\dfrac{1}{2}-\dfrac{3}{4}\)

\(=\dfrac{2}{4}-\dfrac{3}{4}\)

\(=\dfrac{-1}{4}\)

h) \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+20\%\right):\dfrac{7}{3}\)

\(=\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{1}{5}\right):\dfrac{7}{3}\)

\(=\dfrac{7}{5}-\left(\dfrac{11}{20}+\dfrac{4}{20}\right):\dfrac{7}{3}\)

\(=\dfrac{7}{5}-\dfrac{3}{4}:\dfrac{7}{3}\)

\(=\dfrac{7}{5}-\dfrac{9}{28}\)

\(=\dfrac{196}{140}-\dfrac{45}{140}\)

\(=\dfrac{151}{140}\)

i) \(\dfrac{\left(\dfrac{1}{2-0,75}\right).\left(0,2-\dfrac{2}{5}\right)}{\dfrac{5}{9}-1\dfrac{1}{12}}\)

\(=\dfrac{\left(\dfrac{1}{1,25}\right).\left(\dfrac{1}{5}-\dfrac{2}{5}\right)}{\dfrac{5}{9}-\dfrac{13}{12}}\)

\(=\dfrac{\dfrac{1}{1,25}.\dfrac{-1}{5}}{\dfrac{20}{36}-\dfrac{39}{36}}\)

\(=\dfrac{\dfrac{-1}{6,25}}{\dfrac{-19}{36}}\)

k) \(\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{1}{14}}{-1-\dfrac{3}{7}+\dfrac{3}{28}}\)

\(=\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{28}}{-\dfrac{3}{3}-\dfrac{3}{7}+\dfrac{3}{28}}\)

\(=\dfrac{2\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}{\left(-3\right)\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}\)

\(=-\dfrac{2}{3}\)

\(A=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)

\(A=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)

\(A=\left(\dfrac{7}{10}.\dfrac{5}{28}\right).\left(\dfrac{8}{3}.\dfrac{3}{8}\right).20\)

\(A=\dfrac{1}{8}.1.20\)

\(A=\dfrac{20}{8}=\dfrac{5}{2}\)

\(B=\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)

\(B=\left(9\dfrac{3}{8}+7\dfrac{5}{8}\right)+4,03\)

\(B=\left[\left(9+7\right)+\left(\dfrac{3}{8}+\dfrac{5}{8}\right)\right]+4,03\)

\(B=\left(16+1\right)+4,03\)

\(B=17+4,03\)

\(B=21,03\)

\(C=\left(9,75.21\dfrac{3}{7}+\dfrac{39}{4}.18\dfrac{4}{7}\right).\dfrac{15}{78}\)

\(C=\left(\dfrac{39}{4}.\dfrac{150}{7}+\dfrac{39}{4}.\dfrac{130}{7}\right).\dfrac{15}{78}\)

\(C=\dfrac{39}{4}.\left(\dfrac{150}{7}+\dfrac{130}{7}\right).\dfrac{15}{78}\)

\(C=\dfrac{39}{4}.40.\dfrac{15}{78}\)

\(C=390.\dfrac{15}{78}\)

\(C=75\)

\(\frac{x}{5}=\frac23\)

\(x\) = \(\frac23\times5\)

\(x=\frac{10}{3}\)

Vậy \(x=\frac{10}{3}\)

\(\frac{x}{3}-\frac12=\frac15\)

\(\frac{x}{3}\) = \(\frac15\) + \(\frac12\)

\(\frac{x}{3}\) = \(\frac{2}{10}+\frac{5}{10}\)

\(\frac{x}{3}=\frac{7}{10}\)

\(x=\frac{7}{10}\times3\)

\(x=\frac{21}{10}\)

Vậy \(x=\frac{21}{10}\)

\(\frac{x}{5}+\frac12=\frac{6}{10}\)

\(\frac{x}{5}=\frac{6}{10}-\frac12\)

\(\frac{x}{5}=\frac{6}{10}-\frac{5}{10}\)

\(\frac{x}{5}=\frac{1}{10}\)

\(x=\frac{1}{10}\times5\)

\(x=\frac12\)

Vậy \(x=\frac12\)

\(\frac{x+3}{15}\) = \(\frac13\)

\(x+3=\frac13\times15\)

\(x+3=5\)

\(x=5-3\)

\(x=2\)

Vậy \(x=2\)

Đăt A = \(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+......+\frac{1}{7^{100}}\)

\(\Rightarrow7A=1+\frac{1}{7}+\frac{1}{7^2}+.....+\frac{1}{7^{100}}\)

\(\Rightarrow7A-A=1-\frac{1}{7^{100}}\)

\(\Rightarrow6A=1-\frac{1}{7^{100}}\)

\(\Rightarrow A=\frac{1-\frac{1}{7^{100}}}{6}\)