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24 tháng 4 2019

Ta có : 1/1 -1/2 -1/3 +1/2-1/3-1/4+........+1/8-1/9-1/10

            Ta gạch các phân số ở giữa còn lại 1/1-1/10=9/10

24 tháng 4 2019

                       Giải :

    \(=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}-\frac{1}{9\cdot10}\)

\(=\frac{1}{2}-\frac{1}{9\cdot10}=\frac{1}{2}-\frac{1}{90}\)

\(=\frac{45}{90}-\frac{1}{90}=\frac{44}{90}=\frac{22}{45}\)

                                    \(@Cothanhkhe-hoqchac\)

23 tháng 2 2020

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

Cho \(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{8\cdot9\cdot10}\)

Ta có công thức sau: \(\frac{1}{n\cdot\left(n+1\right)}+\frac{1}{\left(n+1\right)\cdot\left(n+2\right)}=\frac{2}{n\cdot\left(n+1\right)\left(n+1\right)}\)

\(\Rightarrow2A=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{8\cdot9\cdot10}\\ =\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}-\frac{1}{9\cdot10}\\ =\frac{1}{1\cdot2}-\frac{1}{9\cdot10}=\frac{22}{45}\)

\(\Rightarrow A=\frac{22}{45}:2=\frac{11}{45}\)

Thay vào phép tính trên ta được:

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

Vậy \(x=\frac{23}{11}\)

12 tháng 10 2019

\(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{8\cdot9\cdot10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{8\cdot9\cdot10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{x}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}-\frac{1}{9\cdot10}\right)=\frac{22}{45}\)

\(\Rightarrow\frac{x}{2}\left(\frac{1}{2}-\frac{1}{90}\right)=\frac{22}{45}\)

\(\Rightarrow\frac{x}{2}\cdot\frac{22}{45}=\frac{22}{45}\)

\(\Rightarrow\frac{x}{2}=1\)

\(\Rightarrow x=2\)

7 tháng 6 2020

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

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

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

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

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

=> \(\frac{11}{45}x=\frac{23}{45}\)

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

Vậy x = 23/11

Ez :))

16 tháng 3 2017

\(pt\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{22}{45}\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{22}{45}\)

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

\(\Leftrightarrow\frac{1}{2}x=1\)

\(\Rightarrow x=2\)

22 tháng 3 2016

Ta có:

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

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

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

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

22 tháng 3 2016

=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

12 tháng 1 2017

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

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

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

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

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

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

5 tháng 4 2017

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

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

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

13 tháng 8 2019

bạn vào câu hỏi tương tự tham khảo nha

13 tháng 8 2019

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

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

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

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

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

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

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

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

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

23 tháng 3 2016

ta có:1/1.2-1/2.3+1/2.3-1/3.4+.....+1/8.9-1/9.10

       =1/1.2-1/9.10=1/2-1/90 

       =45/90-1/90=44/90=22/45

Duyệt nha

23 tháng 3 2016

tk minh 

mk khac tk cho !