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\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{90}\)
\(=\frac{22}{45}\)
Gọi tổng trên là S , ta có :
S = 1/1.2.3 + 1/2.3.4 + ... + 1/8.9.10
S.2 = 2/1.2.3 + 1/2.3.4 + ... + 1/8.9.10
S.2 = 3 -1 /1.2.3 + 4 - 2/2.3.4 + ... + 10 - 8/8.9.10
S.2= 3/1.2.3 - 1/1.2.3 + 4/2.3.4 - 2/2.3.4 + ... + 10/8.9.10 - 8 /8.9.10
S.2 =1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + ... + 1/8.9 - 1/9.10
S.2 = 1/2 - 1/90
S = 1/4 - 1/360
S= 89/360
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{23}{45}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{23}{45}\)
\(\Leftrightarrow\frac{11}{45}x=\frac{23}{45}\)
\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}\)
\(\Rightarrow x=\frac{23}{11}\)
đặt \(A=\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)\)
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\)
\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\)
\(2A=\frac{1}{1.2}-\frac{1}{9.10}=\frac{22}{45}\)
\(A=\frac{22}{45}:2=\frac{11}{45}\)
thay A vào ta được
\(\frac{11}{45}.x=\frac{23}{45}\)
\(x=\frac{23}{45}:\frac{11}{45}=\frac{23}{11}\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{90}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\frac{22}{45}.x=\frac{23}{45}\)
\(\Rightarrow\frac{11}{45}.x=\frac{23}{45}\)
\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}\)
\(\Rightarrow x=\frac{23}{11}\)
Ta có:
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{8.9.10}\right):2.x=\frac{23}{45}\)
\(\Rightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right):2.x=\frac{23}{45}\)
\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right):2.x=\frac{23}{45}\)
\(\Rightarrow\left(\frac{1}{2}-\frac{1}{90}\right):2.x=\frac{23}{45}\)
\(\Rightarrow\frac{11}{45}x=\frac{23}{45}\)
\(\Rightarrow x=\frac{23}{45}\div\frac{11}{45}=\frac{23}{11}\)
Vậy \(x=\frac{23}{11}\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).\frac{1}{2}.x=\frac{23}{45}\)
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{8.9}+\frac{1}{9.10}\right).\frac{1}{2}.x=\frac{23}{45}\)
\(\left(\frac{1}{2}+\frac{1}{6}+....+\frac{1}{72}+\frac{1}{90}\right).\frac{1}{2}.x=\frac{23}{45}\)
\(\left(\frac{1}{2}-\frac{1}{90}\right).\frac{1}{2}.x=\frac{23}{45}\)
\(\frac{22}{45}.\frac{1}{2}x=\frac{23}{45}\)
\(\frac{11}{45}.x=\frac{23}{45}\)
\(x=\frac{23}{45}\div\frac{11}{45}\)
\(x=\frac{23}{11}\)
=> \(x=\frac{23}{11}\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{46}{45}\)
\(=\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x\)
\(=\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{9.10}\right).x\)
\(=\left(\frac{1}{2}-\frac{1}{90}\right).x=\left(\frac{45}{90}-\frac{1}{90}\right)x=\frac{44}{90}.x=\frac{22x}{45}=\frac{46}{45}\)
=> 22x=46
=> x=\(46:22=\frac{23}{11}\)