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

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

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

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

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

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

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

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

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

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

=> \(\dfrac{1}{2}.x=1\)

=> \(x=2\)

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

câu b bài 2:

\(\dfrac{1^2}{1\cdot2}\cdot\dfrac{2^2}{2\cdot3}\cdot\dfrac{3^2}{3\cdot4}\cdot\dfrac{4^2}{4\cdot5}\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\)

\(=\dfrac{1}{5}\)

câu a bài 2:

\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{10\cdot11\cdot12}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}-...-\dfrac{1}{12}\)

\(=1-\dfrac{1}{12}=\dfrac{11}{12}\)