\(B=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)

\(B=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(B=\frac{1}{3}.\left(1-\frac{1}{103}\right)\)

\(B=\frac{1}{3}.\frac{102}{103}\)

\(B=\frac{34}{103}\)

Bài 3: đổi ra phân số rồi tính, đổi:\(1,5=\frac{15}{10};2,5=\frac{25}{10};1\frac{3}{4}=\frac{7}{12}\)(cái này ko giải dùm, đổi ra như thek rồi tính nha)

\(B=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{100.103}\)

\(=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)\)

\(=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(=\frac{1}{3}.\left(1-\frac{1}{103}\right)\)

\(=\frac{1}{3}.\frac{102}{103}\)

\(=\frac{1}{1}.\frac{34}{103}=\frac{34}{103}\)

Thằng an mất dạy :))) 
#Lê_Linh

\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{43.46}\)

    \(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{43}-\frac{1}{46}\)

      \(=1-\frac{1}{46}< 1\)

Vậy \(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{43.46}< 1\)

lick nhật linh là gif

a) \(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)

\(=\frac{5.2^{30}.3^{18}-2^2.2^{27}.3^{20}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)

\(=\frac{2^{29}.3^{18}\left(5.2-3^2\right)}{2^{18}.3^{18}\left(5.3-7.2\right)}\)

\(=\frac{2.1}{1}=2\)

b) \(\frac{0,125-\frac{1}{5}+\frac{1}{7}}{0,375-\frac{3}{5}+\frac{3}{7}}+\frac{\frac{1}{2}+\frac{1}{3}-0,2}{\frac{3}{4}+0,5}-\frac{3}{10}\)

\(=\frac{0,125-\frac{1}{5}+\frac{1}{7}}{3\left(0,125-\frac{1}{5}+\frac{1}{7}\right)}+\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}{\frac{3}{4}+\frac{1}{2}-\frac{3}{10}}\)

\(=\frac{1}{3}+\frac{\frac{30}{60}+\frac{20}{60}-\frac{12}{60}}{\frac{45}{60}+\frac{30}{60}-\frac{9}{60}}\)

\(=\frac{1}{3}+\frac{\frac{19}{30}}{\frac{11}{10}}\)

\(=\frac{1}{3}+\frac{19}{33}=\frac{11}{33}+\frac{19}{33}\)

\(=\frac{30}{33}=\frac{10}{11}\)

Bài nhìn vô muốn xỉu rồi ='((

1. a) \(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{91.94}+\frac{2}{94.97}\)

\(=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{91.94}+\frac{3}{94.97}\right)\)

\(=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}\right)\)

\(=\frac{2}{3}\left(1-\frac{1}{97}\right)=\frac{2}{3}.\frac{96}{97}=\frac{64}{97}\)

b) Bạn tự làm, làm nữa chắc xỉu =((( Khi nào rảnh mình sẽ làm, nếu bạn cần

2 ) 

a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{1005}{2011}\)

\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)

\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{1005}{2011}\)

\(\Leftrightarrow1-\frac{1}{x+2}=\frac{1005}{2011}:2=\frac{1005}{4022}\)

\(\Leftrightarrow\frac{1}{x+2}=1-\frac{1005}{4022}=\frac{3017}{4020+2}\)

\(\Rightarrow x=4020\)

tu ma lam nguoi ta con gap hon min nhieu

Ko bt làm mới đăng chứ

\(a,15\frac{3}{13}-\left(3\frac{4}{7}+8\frac{3}{13}\right)\\ =15\frac{3}{13}-3\frac{4}{7}-8\frac{3}{13}\\ =7\frac{3}{13}-3\frac{4}{7}\\ =\frac{94}{13}-\frac{25}{7}\\ =\frac{94\cdot7-25\cdot13}{13\cdot7}\\ =\frac{333}{91}\)

\(b,\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\\ =7\frac{4}{9}-3\frac{4}{9}+4\frac{7}{11}\\ =4\frac{4}{9}+4\frac{7}{11}\\ =\frac{40}{9}+\frac{51}{11}\\ =\frac{40\cdot11+51\cdot9}{11\cdot9}\\ =\frac{899}{99}\)

\(c,=-\frac{7}{9}\cdot\left(\frac{4}{11}+\frac{7}{11}\right)+\frac{52}{9}\\ =-\frac{7}{9}\cdot1+\frac{52}{9}\\ =\frac{-7+52}{9}\\ =\frac{45}{9}=5\)

\(d,=\frac{50}{100}\cdot\frac{4}{3}\cdot10\cdot\frac{7}{35}\cdot\frac{75}{100}\\ =\frac{50\cdot2\cdot2\cdot2\cdot5\cdot7\cdot25\cdot3}{50\cdot2\cdot3\cdot5\cdot7\cdot25\cdot2\cdot2}=1\)

\(e,=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\\ =1-\frac{1}{43}\\ =\frac{42}{43}\)

a) Ta có: \(15\frac{3}{13}-\left(3\frac{4}{7}+8\frac{3}{13}\right)\)

\(=15+\frac{3}{13}-3-\frac{4}{7}-8-\frac{3}{13}\)

\(=4-\frac{4}{7}=\frac{24}{7}\)

b) Ta có: \(\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\)

\(=7+\frac{4}{9}+4+\frac{7}{11}-3-\frac{4}{9}\)

\(=8+\frac{7}{11}=\frac{95}{11}\)

c) Ta có: \(\frac{-7}{9}\cdot\frac{4}{11}+\frac{-7}{9}\cdot\frac{7}{11}+5\frac{7}{9}\)

\(=\frac{-7}{9}\cdot\frac{4}{11}+\frac{-7}{9}\cdot\frac{7}{11}+\frac{-7}{9}\cdot\frac{-52}{7}\)

\(=\frac{-7}{9}\cdot\left(\frac{4}{11}+\frac{7}{11}-\frac{52}{7}\right)\)

\(=\frac{-7}{9}\cdot\frac{45}{-7}=5\)

d) Ta có: \(50\%\cdot1\frac{1}{3}\cdot10\cdot\frac{7}{35}\cdot0.75\)

\(=\frac{1}{2}\cdot\frac{4}{3}\cdot10\cdot\frac{7}{35}\cdot\frac{3}{4}\)

\(=5\cdot\frac{7}{35}=1\)

e) Ta có: \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)

\(=1-\frac{1}{43}=\frac{43}{43}-\frac{1}{43}\)

\(=\frac{42}{43}\)

\(x+\frac{1}{1.2}+\frac{2}{2.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{5}{11.16}=1\)

\(x+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}=1\)

\(x+1-\frac{1}{16}=1\)

\(x+\frac{15}{16}=1\)

\(x=1-\frac{15}{16}\)

\(x=\frac{1}{16}\)

a, 1/1.2+1/1.3+...+1/99.100

= 1-1/2+1/2-1/3+1/3+...+1/99-1/100

=1-1/100

=99/100

A = 99/100

B = 2014/6051

C = 99

D = 9/38