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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
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 = \(\frac{2}{1}\)
\(\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x=\frac{44}{45}\)
\(\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{44}{45}\)
\(\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{44}{45}\)
\(\frac{22}{45}.x=\frac{44}{45}\)
x =2
Sửa đề: \(A=1.2.3+2.3.4+3.4.5+...+8.9.10\)
\(\Rightarrow4A=1.2.3.4+2.3.4.4+3.4.5.4+...+8.9.10.4\)
\(=2.3.4+2.3.4.\left(5-1\right)+3.4.5.\left(6-2\right)+...+8.9.10.\left(11-7\right)\)
\(=2.3.4+2.3.4.5-2.3.4+3.4.5.6-2.3.4.5+...+8.9.10.11-7.8.9.10\)
\(=8.9.10.11\)
\(=7920\)
\(\Rightarrow A=\dfrac{7920}{4}=1980\)
Có: A=1.2.3+2.3.4+4.5.6+...+8.9.10
4A=1.2.3.4 + 2.3.4.4+...+8.9.10.4
4A=1.2.3.(4-0)+2.3.4.(5-1)+...+8.9.10.(11-7)
4A=1.2.3.4−0.1.2.3+2.3.4.5−1.2.3.4+...+8.9.10.11−7.8.9.1
4A=(1.2.3.4+2.3.4.5+...+8.9.10.11)−(0.1.2.3+1.2.3.4+...+7.8.9.10)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right)x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{8.9.10}\right)x=\frac{22}{45}\)
\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{44}{45}\)
\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{44}{45}\)
\(\Rightarrow\left(\frac{1}{2}-\frac{1}{90}\right)x=\frac{44}{45}\)
\(\Rightarrow\frac{22}{45}.x=\frac{44}{45}\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
đặt tổng trên là A
theo công thức của 7 hàng đẳng thức đáng nhớ ta có:
2A=\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\)
2A=\(\frac{1}{1.2}+\frac{1}{2.3}-\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\)
2A=\(\frac{1}{1.2}-\frac{1}{9.10}\)
2A=\(\frac{1}{2}-\frac{1}{90}\)
2A=\(\frac{22}{45}\)
A=\(\frac{22}{45}:2=\frac{22}{45}.\frac{1}{2}=\frac{11}{45}\)