$x$ ở cuối là sao đây bạn? Nhân riêng với $\frac{1}{8.9.10}$ à?

\(\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{8.9.10}\right).x=\dfrac{23}{45}\)

\(\left[\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{8.9}-\dfrac{1}{9.10}\right)\right].x=\dfrac{23}{45}\)\(\left[\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{9.10}\right)\right].x=\dfrac{23}{45}\)

\(\left(\dfrac{1}{2}.\dfrac{22}{45}\right).x=\dfrac{23}{45}\)

\(\dfrac{11}{45}.x=\dfrac{23}{45}\)

\(x=\dfrac{23}{45}:\dfrac{11}{45}\)

\(x=\dfrac{23}{11}\)

Ta có:

\(\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{8.9.10}\right)x=\dfrac{23}{45}\)

\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{8.9.10}\right)x\) \(=\dfrac{23}{45}\)

\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+...+\dfrac{1}{8.9}-\dfrac{1}{9.10}\right)x\) \(=\dfrac{23}{45}\)

\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{9.10}\right)x=\dfrac{23}{45}\)

\(\Leftrightarrow\dfrac{11}{45}.x=\dfrac{23}{45}\Leftrightarrow x=\dfrac{23}{45}\div\dfrac{11}{45}=\dfrac{23}{11}\)

Vậy \(x=\dfrac{23}{11}\)

\(\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{8.9.10}\right).x=\dfrac{22}{45}\)

=> \(\dfrac{1}{2}.\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{8.9.10}\right).x=\dfrac{22}{45}\)

=> \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-...-\dfrac{1}{9.10}\right).x=\dfrac{22}{45}\)

=> \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{9.10}\right).x=\dfrac{22}{45}\)

=> \(\dfrac{1}{2}.\dfrac{22}{45}.x=\dfrac{22}{45}\)

=> \(\dfrac{1}{2}.x=1\)

=> \(x=2\)

Vậy x = 2

Chúc bạn học tốt !!!

1) [1/(1.2.3)+1/(2.3.4)+...+1/(8.9.10)]x=23/45

[1/(1.2)-1/(2.3)+1/(2.3)-1/(3.4)+...+1/(8.9)-1/(9.10)]x=23/45

[1/(1.2)-1/(9.10)]x=23/45

[1/2-1/90]x=23/45

22/45.x=23/45 => x=23/22

1.

\(\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...........+\dfrac{1}{8.9.10}\right)x=\dfrac{23}{45}\)

\(\Leftrightarrow\left[\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{3.4.5}+............+\dfrac{2}{8.9.10}\right)\right]x=\dfrac{23}{45}\)

\(\Leftrightarrow\left[\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+........+\dfrac{1}{8.9}-\dfrac{1}{9.10}\right)\right]x=\dfrac{23}{45}\)

\(\Leftrightarrow\left[\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{8.9}\right)\right]x=\dfrac{23}{45}\)

\(\Leftrightarrow\left[\dfrac{1}{2}.\dfrac{22}{45}\right]x=\dfrac{23}{45}\)

\(\Leftrightarrow\dfrac{11}{45}.x=\dfrac{23}{45}\)

\(\Leftrightarrow x=\dfrac{23}{11}\)

Vậy \(x=\dfrac{23}{11}\) là giá trị cần tìm

2.

\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+.........+\dfrac{1}{x\left(x+1\right):2}=\dfrac{1998}{2000}\)

\(\Leftrightarrow\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...............+\dfrac{2}{x\left(x+1\right)}=\dfrac{1998}{2000}\)

\(\Leftrightarrow\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...........+\dfrac{2}{x\left(x+1\right)}=\dfrac{1998}{2000}\)

\(\Leftrightarrow2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+.........+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{1998}{2000}\)

\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.........+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{1998}{2000}\)

\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{1998}{2000}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{999}{2000}\)

\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2000}\)

\(\Leftrightarrow x+1=2000\)

\(\Leftrightarrow x=1999\)

Vậy \(x=1999\) là giá trị cần tìm

\(S=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{8.9.10}\)

\(S=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{8.9}-\dfrac{1}{9.10}\)

\(S=\dfrac{1}{1.2}-\dfrac{1}{9.10}\)

\(S=\dfrac{1}{2}-\dfrac{1}{90}=\dfrac{44}{90}\)

\(2S=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{23+24+25}=\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)+\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)+...+\left(\dfrac{1}{23.24}-\dfrac{1}{24.25}\right)\)\(=\dfrac{1}{1.2}-\dfrac{1}{24.25}=\dfrac{299}{600}\) 

Vậy \(S=\dfrac{299}{600}\div2=\dfrac{299}{1200}\)