Sorry đăng làm giwor thì em nó bấm nộp bài mk làm tiếp nhé 

\(E=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+......+\frac{1}{1+2+3+.....+24}\)

\(=\frac{1}{\frac{\left(2-1\right).2}{2}}+\frac{1}{\frac{\left(3-1\right).3}{2}}+.....+\frac{1}{\frac{\left(24-1\right).24}{2}}\)

\(=\frac{1}{\frac{1.2}{2}}+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+.....+\frac{1}{\frac{23.24}{2}}\)

\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+.....+\frac{2}{23.24}\)

\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{23.24}\right)\)

\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{23}-\frac{1}{24}\right)\)

\(=2\left(1-\frac{1}{24}\right)\)

\(=2.\frac{23}{24}=\frac{23}{12}\)

Vậy tỉ số giữa D và E là ; \(\frac{5}{28}:\frac{23}{2}=\frac{5}{322}\)

Ta có : \(D=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+.....+\frac{10}{1400}\)

\(=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+.....+\frac{5}{25.28}\)

\(=\frac{5}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+.....+\frac{3}{25.28}\right)\)

\(=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{25}-\frac{1}{28}\right)\)

\(=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}\right)\)

\(=\frac{5}{3}.\frac{3}{28}=\frac{5}{28}\)

C = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5

C = 1/1 - 1/5

C = 4/5 

\(D=\dfrac{5}{4.7}+\dfrac{5}{7.10}+\dfrac{5}{10.13}+...+\dfrac{5}{25.28}\)

\(=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{28}\right)=\dfrac{5}{3}.\dfrac{6}{28}=\dfrac{5}{14}\)

\(E=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{24.25}=2\left(\dfrac{1}{2}-\dfrac{1}{25}\right)=\dfrac{2.23}{50}=\dfrac{23}{25}.\)

\(\dfrac{D}{E}=\dfrac{5}{24}.\dfrac{25}{23}=\dfrac{125}{552}.\)

sai ròi kìa

mọi người ơi giúp em vs ạ , e đang rất cần 

\(1+2+...+n=\dfrac{\left(\dfrac{n-1}{1}+1\right).\left(n+1\right)}{2}=\dfrac{n\left(n+1\right)}{2}\)

\(M=\dfrac{3}{1+2}+\dfrac{3}{1+2+3}+...+\dfrac{3}{1+2+...+2022}\)

\(=3\left(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+...+2022}\right)\)

\(=3\left(\dfrac{1}{\dfrac{2.\left(2+1\right)}{2}}+\dfrac{1}{\dfrac{3.\left(3+1\right)}{2}}+...+\dfrac{1}{\dfrac{2022.\left(2022+1\right)}{2}}\right)\)

\(=3\left(\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{2022.2023}\right)\)

\(=3.2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\right)\)

\(=6.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)

\(=6.\left(\dfrac{1}{2}-\dfrac{1}{2023}\right)\)

\(=6.\dfrac{2021}{4046}=3.\dfrac{2021}{2023}=\dfrac{6063}{2023}=\dfrac{18189}{6069}\)

\(\dfrac{10}{3}=\dfrac{20230}{6069}>\dfrac{18189}{6069}=M\)

a) \(\frac{-2}{3}x+\frac{1}{5}=\frac{1}{10}\)

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

\(\Leftrightarrow\frac{-2}{3}x=\frac{-1}{10}\)

\(\Leftrightarrow x=\frac{-1}{10}\div\frac{-2}{3}\)

\(\Leftrightarrow x=\frac{3}{20}\)

ko có tên à

\(E=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{10.11.12}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-...-\frac{1}{10.11}+\frac{1}{10.11}-\frac{1}{11.12}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{11.12}\right)=\frac{1}{2}.\frac{65}{132}=\frac{65}{264}\)

Ta có : 

\(E=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{10.11.12}\)

\(2E=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{10.11.12}\)

\(2E=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{10.11}-\frac{1}{11.12}\)

\(2E=\frac{1}{1.2}-\frac{1}{10.11}\)

\(2E=\frac{1}{2}-\frac{1}{110}\)

\(2E=\frac{27}{55}\)

\(E=\frac{27}{55}:2\)

\(E=\frac{27}{55}.\frac{1}{2}\)

\(E=\frac{27}{110}\)

Vậy \(E=\frac{27}{110}\)

Chúc bạn học tốt ~ 

1)C= 1/5+1/10+1/20+1/40+...+1/1280

\(=\frac{1}{5}\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)\)

Đặt cái trong ngoặc là A ta có:\(2A=2+1+...+\frac{1}{2^7}\)

\(2A-A=\left(2+1+...+\frac{1}{2^7}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^8}\right)\)

\(A=2-\frac{1}{2^8}\).Thay A vào ta được:\(C=\frac{1}{5}\left(2-\frac{1}{2^8}\right)=\frac{1}{5}\cdot\frac{511}{256}=\frac{511}{1280}\)

2)D= 2/1*3+2/3*5+2/5*10+2/7*9+2/9*11+2/11*18+2/13*15

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\)

\(=1-\frac{1}{15}\)

\(=\frac{14}{15}\)

3)E= 4/3*7+4/7*11+4/11*15+4/15*19+4/19*23+4/23*27

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}\)

\(=\frac{1}{3}-\frac{1}{27}\)

\(=\frac{8}{27}\)

4)G= 1/2+1/6+1/12+1/20+1/30+1/42+...+1/110

\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)

\(=1-\frac{1}{11}\)

\(=\frac{10}{11}\)

5)H= 3/1*2+3/2*3+3/3*4+3/4*5+...+3/9*10

\(=3\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=3\left(1-\frac{1}{10}\right)\)

\(=3\times\frac{9}{10}\)

\(=\frac{27}{10}\).Lần sau bạn đăng ít một thôi nhé 

Nguyễn Trà My

Phần a)

\(3\times\left(\frac{1}{2}-x\right)+\frac{1}{3}=\frac{7}{6}-x\)

\(32-3x+13=76-x\)

\(116-3x=76-x\)

\(116-76=3x-x\)

\(46=2x\)

\(x=46\div2\)

\(x=13\)

a)  \(3.\left(\frac{1}{2}-x\right)+\frac{1}{3}=\frac{7}{6}-x\)

\(3.\left(\frac{1}{2}-x\right)+x=\frac{7}{6}-\frac{1}{3}\)

\(\Rightarrow\frac{3}{2}-3x+x=\frac{5}{6}\)

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

\(2x=-\frac{2}{3}\)

\(x=-\frac{2}{3}:2\)

\(x=-\frac{1}{3}\)

\(E=1-\dfrac{1}{2^2}-\dfrac{1}{2^3}-\dfrac{1}{2^4}-...-\dfrac{1}{2^{10}}\)

\(E=1-\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{10}}\right)\)

Đặt \(S=\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{10}}\)

\(2S=2\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{10}}\right)\)

\(2S=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}\)

\(2S-S=\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\right)\)

\(S=\dfrac{1}{2}-\dfrac{1}{2^{10}}\). Khi đó \(E=1-\left(\dfrac{1}{2}-\dfrac{1}{2^{10}}\right)=1-\dfrac{1}{2}+\dfrac{1}{2^{10}}=\dfrac{513}{1024}\)