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3 tháng 2 2019

1/5.8+1/8.11+....+1/x(x+3)=1/6

=> 3/5.8 + 3/8.11 +.....+ 3/x(x+3)=1/2

=> 1/5-1/x+3=1/2

=> 1/x+3=1/5-1/2=-3/10=1/-10/3

=>x+3=-10/3=>x=-19/3

3 tháng 2 2019

                   \(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{1}{6}\)

                 \(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\cdot\left(x+3\right)}=\frac{1}{2}\)

\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{x}-\frac{1}{x+3}=\frac{1}{2}\)

                                                              \(\frac{1}{5}-\frac{1}{x+3}=\frac{1}{2}\)

                                                                          \(\frac{1}{x+3}=\frac{-3}{10}\)

\(\Rightarrow\left(-3\right)\left(x+3\right)=10\)

\(\Rightarrow x+3=\frac{-10}{3}\)

\(\Rightarrow x=\frac{-19}{3}\)

Vậy \(x=\frac{-19}{3}\)

19 tháng 8 2020

\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{1}{2}\)

\(\Leftrightarrow\frac{1}{x+3}=-\frac{3}{10}\)

\(\Leftrightarrow1\cdot10=-3\left(x+3\right)\)

\(\Leftrightarrow10=-3x-9\)

\(\Leftrightarrow10+9=-3x\)

\(\Leftrightarrow19=-3x\)

\(\Leftrightarrow x=-\frac{19}{3}\)

Đề sai à -.- 

19 tháng 8 2020

\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{1}{6}\)

=> \(\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{1}{6}\)

=> \(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{1}{6}:\frac{1}{3}\)

=> \(\frac{1}{5}-\frac{1}{x+3}=\frac{1}{6}\cdot3=\frac{1}{2}\)

=> \(\frac{1}{x+3}=\frac{1}{5}-\frac{1}{2}=-\frac{3}{10}\)

=> \(10=-3\left(x+3\right)\)

=> 10 = -9x - 9

=> 10 + 9x + 9 = 0

=> 19 + 9x = 0

=> 9x = -19

=> x = -19/9

2 tháng 5 2022

giúp mik đi ạ

6 tháng 5 2019

Đặt 1/5.8 + 1/8.11 +...+ 1 /x (x+3) = A

3A = 3/5.8 + 3/8.11 +...+ 3/x (x+3)

3A = 1/5 - 1/8 + 1/8 - 1/11 +...+ 1/x - 1/x+3

3A = 1/5 - 1/x + 3 

3A = ( 3+x)-5/5x +15

A =[ ( 3+ x ) - 5 / 5x + 15 ] : 3

A = x + ( - 2 ) / 5x + 15

   Ta có :

A + 27/480

= x + ( - 2 ) / 5x + 15

=> x + ( - 2 ) = 27

=> 5x + 15 = 480

          * Làm nốt *

                #Louis

6 tháng 5 2019

\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{27}{480}\)

\(=\frac{1}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{27}{480}\)

\(=\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{27}{480}\)

\(=\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{27}{480}\)

\(=\frac{1}{5}-\frac{1}{x+3}=\frac{27}{480}.3\)

\(=\frac{1}{5}-\frac{1}{x+3}=\frac{81}{480}\)

\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{81}{480}=\frac{15}{480}=\frac{1}{32}\)

\(\Rightarrow x+3=32\)

\(\Rightarrow x=32-3=29\)

7 tháng 3 2017

dơn giản như đan rổ

16 tháng 4 2021

đề thiếu nhé

16 tháng 4 2021

Ta có: \(\dfrac{k}{x.\left(x+k\right)}=\dfrac{x+k-x}{x.\left(x+k\right)}=\dfrac{1}{x}-\dfrac{1}{x+k}\)

nên áp dụng ta có:

\(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{x.\left(x+3\right)}\)

\(=\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\)

\(=\dfrac{1}{5}-\dfrac{1}{x+3}\)

Nên $\dfrac{1}{3}.\left(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{x.\left(x+3\right)}\right)=\dfrac{1}{3}.(\dfrac{1}{5}-\dfrac{1}{x+3})$
Đến đây là làm được rồi nha

22 tháng 9 2016

a) \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+....+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)

\(=3.\left(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}\right)=\frac{101}{1540}.3\)

\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x.3}=\frac{303}{1540}\)

\(=\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

\(=\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)

\(=\frac{1}{x+3}=\frac{1}{308}\)

\(x+3=308\)

\(\Rightarrow x=305\)