\(x\times\left(\frac{2015}{8\times9}+\frac{1925}{9\times10}+\frac{1795}{10\times11}+\frac{1629}{11\times12}+6\right)=\frac{1}{24}\)

=>    \(x\times\left(\frac{2015}{72}+\frac{1925}{90}+\frac{1795}{110}+\frac{1629}{132}+6\right)=\frac{1}{24}\)

=>   \(x\times84\frac{3}{88}=\frac{1}{24}\Rightarrow x=\frac{1}{24}:\frac{7395}{88}=\frac{1}{24}\times\frac{88}{7395}=\frac{88}{177480}=\frac{11}{22185}\)

\(2\left(\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1}{9}\)

\(\Leftrightarrow2\left(\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1}{9}\)

\(\Leftrightarrow2\left(\frac{1}{9}-\frac{1}{x+1}\right)=\frac{1}{9}\)

\(\Leftrightarrow\frac{1}{9}-\frac{1}{x+1}=\frac{1}{9}\div2\)

\(\Leftrightarrow\frac{1}{9}-\frac{1}{x+1}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{9}-\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{18}\)

\(\Leftrightarrow x+1=18\)

\(\Leftrightarrow x=18-1\)

\(\Leftrightarrow x=17\)

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

\(\Leftrightarrow\left|x\right|=\frac{5}{3}+\frac{3}{4}\)

\(\Leftrightarrow\left|x\right|=\frac{20}{12}+\frac{9}{12}\)

\(\Leftrightarrow\left|x\right|=\frac{29}{12}\)

\(\Leftrightarrow x=\pm\frac{29}{12}\)

Lưu ý : . là dấu nhân

ai làm đúng , nhanh mình k

ko ghi lại đề

\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{210}-\frac{1}{211}+\frac{1}{211}-\frac{1}{212}\)

\(=1-\frac{1}{212}\)

\(=\frac{211}{212}\)

\(\frac{1}{8.9}+\frac{1}{9.10}+...+\frac{1}{211.212}\)

= \(\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+...+\frac{1}{211}-\frac{1}{212}\)

= \(\frac{1}{8}-\frac{1}{212}\)

= \(\frac{51}{424}\)

A=\(\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{13}-\frac{1}{14}\)=\(\frac{1}{7}-\frac{1}{14}\)=\(\frac{1}{14}\)

B=0

\(\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}\)

\(=\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}\)

\(=\frac{1}{7}-\frac{1}{14}=\frac{1}{14}\)

\(\frac{7}{3.4}-\frac{9}{4.5}+\frac{11}{5.6}-\frac{13}{6.7}+\frac{15}{7.8}-\frac{17}{8.9}-\frac{19}{9.10}+\frac{21}{10.11}\)

\(=\frac{3+4}{3.4}-\frac{4+5}{4.5}+\frac{5+6}{5.6}-\frac{6+7}{6.7}+\frac{7+8}{7.8}-\frac{8+9}{8.9}-\frac{9+10}{9.10}+\frac{10+11}{10.11}\)

\(=\frac{1}{3}+\frac{1}{4}-\frac{1}{4}-\frac{1}{5}+\frac{1}{5}+\frac{1}{6}-\frac{1}{6}-\frac{1}{7}+\frac{1}{7}+\frac{1}{8}-\frac{1}{8}-\frac{1}{9}+\frac{1}{9}+\frac{1}{10}-\frac{1}{10}-\frac{1}{11}\)

\(=\frac{1}{3}-\frac{1}{11}=\frac{8}{33}\)

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

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

=> \(\frac{9}{10}.\left(x-1\right)=x-\frac{1}{3}\)

=> \(\frac{9x}{10}-\frac{9}{10}=\frac{3x-1}{3}\)

=> \(\frac{27x}{30}-\frac{27}{30}=\frac{10.\left(3x-1\right)}{30}\)

=> 27x - 27 = 30x - 10

=> 27x - 30x = -10 + 27

=> -3x = 17

=> x = -17/3.