Đặt \(A=\frac{7}{1\cdot5}+\frac{7}{5\cdot9}+\frac{7}{9\cdot13}+\frac{7}{13\cdot17}+\frac{7}{17\cdot21}\)

\(\frac{4}{7}A=\frac{4}{7}\left(\frac{7}{1\cdot5}+\frac{7}{5\cdot9}+\frac{7}{9\cdot13}+\frac{7}{13\cdot17}+\frac{7}{17\cdot21}\right)\)

\(\frac{4}{7}A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}\)

\(\frac{4}{7}A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}\)

\(\frac{4}{7}A=1-\frac{1}{21}\)

\(\frac{4}{7}A=\frac{20}{21}\)

\(A=\frac{20}{21}:\frac{4}{7}=\frac{20}{21}\cdot\frac{7}{4}=\frac{5}{3}\)

Đặt biểu thức trên là A

\(\frac{4xA}{7}=\frac{4}{1x5}+\frac{4}{5x9}+\frac{4}{9x13}+\frac{4}{13x17}+\frac{4}{17x21}\)

\(\frac{4xA}{7}=\frac{5-1}{1x5}+\frac{9-5}{5x9}+\frac{13-9}{9x13}+\frac{17-13}{13x17}+\frac{21-17}{17x21}\)

\(\frac{4xA}{7}=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}\)

\(\frac{4xA}{7}=1-\frac{1}{21}=\frac{20}{21}\Rightarrow A=\frac{20}{21}.\frac{7}{4}=\frac{5}{3}\)

Toán quá dễ. Tự túc là hạnh phúc mọi nhà bn nhé !

\(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+\frac{3}{4.5}+\frac{3}{5.6}+...+\frac{3}{9.10}+\frac{77}{2.9}+\frac{77}{9.16}+\frac{77}{16.23}+...+\frac{77}{93.100}\)

Gọi \(\left(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+......+\frac{3}{9.10}\right)\)là \(A\); \(\left(\frac{77}{2.9}+\frac{77}{9.16}+\frac{77}{16.23}+...+\frac{77}{93.100}\right)\)là B . Ta có : 

\(A=\frac{3}{1}.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)

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

\(A=\frac{3}{1}\cdot\frac{9}{10}=\frac{27}{10}\)

\(B=\frac{77}{7}\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{6}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+....+\frac{1}{93}-\frac{1}{100}\right)\)

\(B=\frac{77}{7}\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(B=\frac{77}{7}\cdot\frac{49}{100}=\frac{539}{100}\)

\(\Rightarrow\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+\frac{3}{4.5}+...+\frac{3}{9.10}+\frac{77}{2.9}+\frac{77}{9.16}+\frac{77}{16.23}+...+\frac{77}{93.100}=\frac{27}{10}+\frac{539}{100}=\frac{809}{100}\)

\(\frac{4}{3.6}+\frac{4}{6.9}+\frac{4}{9.12}+\frac{4}{12.15}\)

\(=\frac{4}{3}\cdot\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\right)\)

\(=\frac{4}{3}\cdot\left(\frac{1}{3}-\frac{1}{15}\right)\)

\(=\frac{4}{3}\cdot\frac{4}{15}=\frac{16}{45}\)

\(\frac{7}{1.5}+\frac{7}{5.9}+\frac{7}{9.13}+\frac{7}{13.17}+\frac{7}{17.21}\)

\(=\frac{7}{4}\cdot\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}\right)\)

\(=\frac{7}{4}\cdot\left(\frac{1}{1}-\frac{1}{21}\right)\)

\(=\frac{7}{4}\cdot\frac{20}{21}=\frac{5}{3}\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{110}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{10.11}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{10}-\frac{1}{11}\)

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

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

Sủa đề : \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)

\(=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)

\(=\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)

\(=\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(=\frac{1}{3}\cdot\frac{9}{20}=\frac{3}{20}\)

Ta có:\(\frac{1}{3x4}+\frac{2}{4x6}+\frac{3}{6x9}+\frac{4}{9x13}+\frac{5}{13x18}\)

\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}\)

\(=\frac{1}{3}-\frac{1}{18}=\frac{6}{18}-\frac{1}{18}=\frac{5}{18}\)

đặt \(A=\frac{1}{3.4}+\frac{2}{4.6}+\frac{3}{6.9}+\frac{4}{9.13}+\frac{5}{13.18}\)=\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}=\frac{1}{3}-\frac{1}{18}=\frac{5}{18}\)

=>A=5/18

vậy........

\(\frac{1}{3.4}+\frac{1}{4.6}+\frac{1}{6.9}+\frac{1}{9.13}+\frac{5}{13.18}\)

\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}\)

\(=\frac{1}{3}-\frac{1}{18}\)

\(=\frac{5}{18}\)

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)

\(=\frac{1\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}{4.\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)

\(=\frac{1}{4}+\frac{3}{4}=1\)

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)

\(=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)

\(=\frac{1}{4}+\frac{3}{4}\)

=1

Ta có:

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left[\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right]}=\frac{1}{4}\)

Lại có:

\(\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}=\frac{3\left[\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right]}{4\left[\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right]}=\frac{3}{4}\)

Vậy: 

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}=\frac{1}{4}+\frac{3}{4}=1\)

nhanh lên các bạn nhé !

\(4\cdot\left(x-12\right)=2x+164\)

\(\Leftrightarrow4x-48=2x+164\)

\(\Leftrightarrow4x-2x=164+48\)

\(\Leftrightarrow2x=212\Rightarrow x=106\)

Sửa đề : \(\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{401.405}+x=2\frac{1}{405}\)

\(\Leftrightarrow\frac{1}{4}\cdot\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{401}-\frac{1}{405}\right)+x=\frac{811}{405}\)

\(\Leftrightarrow\frac{1}{4}\cdot\left(1-\frac{1}{405}\right)+x=\frac{811}{405}\)

\(\Leftrightarrow\frac{1}{4}\cdot\frac{404}{405}+x=\frac{811}{405}\)

\(\Leftrightarrow\frac{101}{405}+x=\frac{811}{405}\Rightarrow x=\frac{142}{81}\)

   4x(X-12)=2xX+164

2x2x(X-12)=2xX+2x82

  2x(2X-24)=2x(X+82)

        2X-24=X+82

            2X=X+82+24

            2X=X+106

             X=106

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)

\(=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)

\(=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1\)