Ta có:

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{90}\right)\)

=\(\frac{11}{45}\)

=1/2.(1/1*2-1/2*3+1/2*3-1/3*4+...+1/8*9-1/9*10)

=1/2.(1/1*2-1/9*10)

=1/2.44/90

=22/90

nha

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{23}{45}\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{23}{45}\)

\(\Leftrightarrow\frac{11}{45}x=\frac{23}{45}\)

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

\(\Rightarrow x=\frac{23}{11}\)

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

\(\frac{11}{45}.x=\frac{23}{45}\)

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

bạn vào câu hỏi tương tự tham khảo nha

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{90}\right).x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}.\frac{22}{45}.x=\frac{23}{45}\)

\(\Rightarrow\frac{11}{45}.x=\frac{23}{45}\)

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

\(\Rightarrow x=\frac{23}{11}\)

=>(2/1.2.3+2/2.3.4+....+2/8.9.10).x=22/45

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

=>(1/1.2-1/9.10).x=22/45

=>22/45.x=44/45

=>x=2

4+2^2+2^3+....+2^20=2^n

=>2^2+2^2+2^3+....+2^20=2^n

đặt 2^2+2^3+....+2^20

=>2A-A=2^21-2^2

khi đó A=2^2+2^21-2^2=2^21=2^n

=>n=21

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 \(x=2\)

2

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\\ =\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2450}\right)\)

\(=\frac{1}{2}.\frac{612}{1225}\\ =\frac{306}{1225}\)(mà đây là toán 6 mà :V)