Lời giải:

Đặt $A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{8.9.10}$

$2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{8.9.10}$

$=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{10-8}{8.9.10}$

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

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

$A=\frac{11}{45}$

$Ax=\frac{11}{45}x=\frac{22}{45}$

$x=\frac{22}{45}: \frac{11}{45}=2$

\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\cdot x=\frac{22}{45}\)

\(\Rightarrow1-\frac{1}{10}.x=\frac{22}{45}\)

\(\Rightarrow\frac{9}{10}.x=\frac{22}{45}\)

\(\Rightarrow x=\frac{22}{45}:\frac{9}{10}\)

\(\Rightarrow x=\frac{22}{45}.\frac{10}{9}=\frac{22.10}{45.9}=\frac{44}{81}\)

=>x=\(\frac{44}{81}\)

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\)

\(A=\frac{1}{2}\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{10-8}{8.9.10}\right)\)

\(A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)

\(A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{11}{45}\).

Phương trình tương đương với: 

\(\frac{11}{45}x=\frac{22}{45}\Leftrightarrow x=2\).

nhấn vào đúng 0 sẽ ra đáp án

Đặt A=11.2.3+12.3.4+....+18.9.10A=11.2.3+12.3.4+....+18.9.10

2A=21.2.3+22.3.4+....+28.9.102A=21.2.3+22.3.4+....+28.9.10

=3−11.2.3+4−22.3.4+...+10−88.9.10=3−11.2.3+4−22.3.4+...+10−88.9.10

=11.2−12.3+12.3−13.4+...+18.9−19.10=11.2−12.3+12.3−13.4+...+18.9−19.10

=11.2−19.10=2245=11.2−19.10=2245

A=1145A=1145

Ax=1145x=2245Ax=1145x=2245

x=2245:1145=2

(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

= 1/ 2 ( 1/1.2 - 1/2.3 +1/2.3-1/3.4+...+1/8.9-1/9.10 ) x = 22/45

= 1/2 .(1/2-1/90)x = 22/45

= 1/2 . 22/45. x=22/45

x = 22/45:( 22/45 .1/2)

x =2 

duyệt