Sửa đề : \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\Leftrightarrow\)\(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(\Leftrightarrow\)\(\frac{1}{6}-\frac{1}{x+1}=\frac{2}{9}.\frac{1}{2}\)

\(\Leftrightarrow\)\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

\(\Leftrightarrow\)\(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)

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

\(\Leftrightarrow\)\(x+1=18\)

\(\Leftrightarrow\)\(x=18-1\)

\(\Leftrightarrow\)\(x=17\)

Vậy \(x=17\)

Chúc bạn học tốt ~ 

\(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)

\(\Leftrightarrow\)\(x+10\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{53.55}\right)=\frac{3}{11}\)

\(\Leftrightarrow\)\(x+10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)

\(\Leftrightarrow\)\(x+10\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)

\(\Leftrightarrow\)\(x+10.\frac{4}{55}=\frac{3}{11}\)

\(\Leftrightarrow\)\(x+\frac{40}{55}=\frac{3}{11}\)

\(\Leftrightarrow\)\(x=\frac{3}{11}-\frac{40}{55}\)

\(\Leftrightarrow\)\(x=\frac{-5}{11}\)

Vậy \(x=\frac{-5}{11}\)

Chúc bạn học tốt ~ 

\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{9}\)

\(\Rightarrow\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{9}\)

\(\Rightarrow\frac{2}{6\cdot7}+\frac{2}{7\cdot8}+\frac{2}{8\cdot9}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{9}\)

\(\Rightarrow2\left(\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)

\(\Rightarrow2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(\Rightarrow2\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(\Rightarrow\frac{1}{3}-\frac{2}{x+1}=\frac{2}{9}\)

\(\Rightarrow\frac{2}{x+1}=\frac{1}{9}\)

=> 2.9 = x + 1

=> x + 1 = 18

=> x = 17

\(\frac{33}{2}+\frac{33}{6}+\frac{33}{18}+\frac{33}{54}+\frac{33}{162}+\frac{33}{486}\)

\(=\frac{33.3+33.3+33.3+33.3+33.3}{486}\)

\(=\frac{99.5}{486}\)

\(=\frac{495}{486}\)

Gọi \(A=\frac{33}{2}+\frac{33}{6}+...+\frac{33}{486}\)

\(A=33.\left[\left(\frac{1}{1.2}+\frac{1}{2.3}\right)+\left(\frac{1}{3.6}+\frac{1}{6.9}\right)\left(\frac{1}{9.18}+\frac{1}{18.27}\right)\right]\)

\(A=33.\left[\frac{2}{3}+\frac{2}{9}+\frac{2}{27}\right]\)

\(A=66.\left[\frac{9}{27}+\frac{3}{27}+\frac{1}{27}\right]\)

\(A=66.\frac{13}{27}\)

\(A=\frac{286}{9}\)

sai hay đúng cx ko biết nha

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)

\(=1+\frac{2}{3}-\frac{2}{3}+\frac{2}{5}-\frac{2}{5}+....+\frac{2}{15}\)

\(=1+\frac{2}{15}\)

\(=\frac{17}{15}\)

\(164:\left\{192-\left[2^3\cdot15-\left(40-37\right)^2\right]+2021^0\right\}\)

\(=164:\left\{192-120+9+1\right\}\)

\(=164:82=2\)

\(\frac{2^{10}\cdot55+2^{10}\cdot26}{2^8\cdot27}\)

\(=\frac{2^{10}\left(55+26\right)}{2^8\cdot27}\)

\(=\frac{2^{10}\cdot\text{81}}{27}\)

\(=2^2\cdot3\)

\(=12\)

\(\frac{2^{10}.55+2^{10}.26}{2^8.27}=\frac{2^{10}.\left(55+26\right)}{2^8.27}\)

                                    \(=\frac{2^2.81}{27}\)

                                      \(=2^2.3\)

                                      \(=12\)

\(\frac{2^{10}.55+2^{10}.26}{2^8.27}\)\(=\frac{2^{10}.\left(55+26\right)}{2^8.27}\)\(=\frac{2^{10}.81}{2^8.27}\)

\(=\frac{2^{10}.3^4}{2^8.3^3}\)\(=\frac{2^8.2^2.3^3.3}{2^8.3^3}=2^2.3=12\)

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+....+\frac{1}{97.100}=\frac{0,33.x}{2009}\)

\(\Leftrightarrow\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{97}-\frac{1}{100}\right)=\frac{0,33.x}{2009}\)

\(\Leftrightarrow\frac{1}{3}\cdot\left(1-\frac{1}{100}\right)=\frac{0,33.x}{2009}\)

\(\Leftrightarrow\frac{1}{3}\cdot\frac{99}{100}=\frac{0,33.x}{2009}\)

\(\Leftrightarrow\frac{33}{100}=\frac{0,33.x}{2009}\)

\(\Leftrightarrow x=\frac{0,33\times100}{0,33}=100\)

toán 1 sao?

\(\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+...+\frac{5}{53\cdot55}\)

\(=\frac{3}{2}\cdot\left(\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{53\cdot55}\right)\)

\(=\frac{3}{2}\cdot\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{53}-\frac{1}{55}\right)\)

\(=\frac{3}{2}\cdot\left(\frac{1}{5}-\frac{1}{55}\right)\)

\(=\frac{3}{2}\cdot\left(\frac{11}{55}-\frac{1}{55}\right)\)

\(=\frac{3}{2}\cdot\frac{10}{55}\)

\(=\frac{3}{2}\cdot\frac{2}{11}\)

\(=\frac{3}{11}\)

đây mà là lớp 1 ak