Lời giải:

Gọi tổng trong ngoặc là $A$
$2A=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+....+\frac{10-8}{8.9.10}$

$=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}$

$=\frac{1}{1.2}-\frac{1}{9.10}=\frac{1}{2}-\frac{1}{90}=\frac{22}{45}$

Vậy $\frac{22}{45}x=\frac{23}{45}$

$\Rightarrow x=\frac{23}{45}: \frac{22}{45}=\frac{23}{22}$

$x$ 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 \(x=\dfrac{23}{11}\)

\(linh_1=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}\)

\(linh_1=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}\right)\)

\(linh_1=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{4.5}\right)\)

\(linh_1=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{20}\right)=\dfrac{1}{2}.\dfrac{9}{20}=\dfrac{9}{40}\)

\(linh_2=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{8.9.10}\)

\(linh_2=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{8.9}-\dfrac{1}{9.10}\right)\)\(linh_2=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{9.10}\right)\)

\(linh_2=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{90}\right)=\dfrac{1}{2}.\dfrac{22}{45}=\dfrac{11}{45}\)

a/ \(G=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}\)

\(\Leftrightarrow2G=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}\)

\(\Leftrightarrow2G=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}\)

\(\Leftrightarrow2G=\dfrac{1}{1.2}-\dfrac{1}{4.5}\)

\(\Leftrightarrow2G=\dfrac{1}{2}-\dfrac{1}{20}\)

\(\Leftrightarrow2G=\dfrac{9}{20}\)

\(\Leftrightarrow G=\dfrac{9}{40}\)

b/ \(H=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+.....+\dfrac{1}{8.9.10}\)

\(\Leftrightarrow2H=\dfrac{2}{1.2.3}+\dfrac{2}{3.4.5}+.....+\dfrac{2}{8.9.10}\)

\(\Leftrightarrow2H=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+.....+\dfrac{1}{8.9}-\dfrac{1}{9.10}\)

\(\Leftrightarrow2H=\dfrac{1}{1.2}-\dfrac{1}{9.10}\)

\(\Leftrightarrow2H=\dfrac{1}{2}-\dfrac{1}{90}\)

\(\Leftrightarrow2H=\dfrac{22}{45}\)

\(\Leftrightarrow H=\dfrac{22}{90}\)

(1/1.2.3+1/2.3.4+...+1/8.9.10).x=22/45

<=>(2/1.2.3+2/2.3.4+...+2/8.9.10).x=44/45

<=>(1/1.2-1/2.3+1/2.3-1/3.4+...+1/8.9-1/9.10).x=44/45

<=>(1/1.2-1/9.10).x=44/45

<=>22/45.x=44/45

<=>x=2

vậy x=2 

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

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

\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+............+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{22}{45}\)

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

\(\Rightarrow\frac{1}{2}.\frac{22}{45}.x=\frac{22}{45}\)

\(\Rightarrow\frac{11}{45}.x=\frac{22}{45}\)

\(\Rightarrow x=\frac{22}{45}:\frac{11}{45}\)

\(\Rightarrow x=2\)

Mình đồng ý với cách làm của cả hai bạn , nhưng mình thấy bạn Nguyễn Hung Phát làm rõ ràng hơn!!!

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