Lời giải:

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

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

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

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

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

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

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

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

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

\(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{8\cdot9\cdot10}\right)x=\frac{23}{45}\)

=> \(\left[\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{8\cdot9\cdot10}\right)\right]x=\frac{23}{45}\)

=>\(\left[\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}-\frac{1}{9\cdot10}\right)\right]x=\frac{23}{45}\)

=> \(\left[\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9\cdot10}\right)\right]x=\frac{23}{45}\)

=> \(\left[\frac{1}{2}\cdot\frac{22}{45}\right]x=\frac{23}{45}\)

=> \(\frac{11}{45}x=\frac{23}{45}\)

=> \(x=\frac{23}{45}:\frac{11}{45}=\frac{23}{45}\cdot\frac{45}{11}=\frac{23}{11}\)

Vậy x = 23/11

Ez :))