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Lời giải:
Đặt $A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{8.9.10}$
$2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{8.9.10}$
$=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{10-8}{8.9.10}$
$=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}$
$=\frac{1}{1.2}-\frac{1}{9.10}=\frac{22}{45}$
$A=\frac{11}{45}$
$Ax=\frac{11}{45}x=\frac{22}{45}$
$x=\frac{22}{45}: \frac{11}{45}=2$
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
\(\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x=\frac{44}{45}\)
\(\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{44}{45}\)
\(\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{44}{45}\)
\(\frac{22}{45}.x=\frac{44}{45}\)
x =2
Đặt B = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{8.9.10}\)
=> 2B = \(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+....+\frac{2}{8.9.10}\)
=> 2B = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{8.9}-\frac{1}{9.10}\)
=> 2B = \(\frac{1}{1.2}-\frac{1}{9.10}\)
2B = \(\frac{22}{45}\)
B = \(\frac{22}{45}:2\)
=> B = \(\frac{11}{45}\)
Ta có : \(\frac{11}{45}.x=\frac{22}{45}\)
=> x = \(\frac{22}{45}:\frac{11}{45}\)
=> x = \(\frac{2}{1}\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right)x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{8.9.10}\right)x=\frac{22}{45}\)
\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{44}{45}\)
\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{44}{45}\)
\(\Rightarrow\left(\frac{1}{2}-\frac{1}{90}\right)x=\frac{44}{45}\)
\(\Rightarrow\frac{22}{45}.x=\frac{44}{45}\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{22}{45}\)
\(\Leftrightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+..+\frac{2}{8.9.10}\right).x=\frac{44}{45}\)
\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{44}{45}\)
\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{44}{45}\Leftrightarrow\frac{22}{45}.x=\frac{44}{45}\Leftrightarrow x=2\)
Vậy x=2