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

=)\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)

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

Suy ra \(\frac{1}{5}-\frac{1}{x+3}\)= \(\frac{303}{1540}\)=)\(\frac{1}{x+3}=\frac{1}{305}\)=)   \(x+3=305\)=) \(x=302\)

A ) \(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+.....+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}.\)

=\(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)\)=101/1540

=\(\frac{101}{1540}:\frac{1}{3}=\frac{1}{5}-\frac{1}{x+3}\)

=tới đó bn tự tính nhé

Bấm vào câu hỏi nha , mk có sửa chút !

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

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

\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-...-\frac{1}{x+3}=\frac{101}{1540}\)

\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{101}{1540}=\frac{207}{1540}\)

\(\frac{1}{x+3}=\frac{207}{1540}\Leftrightarrow207\left(x+3\right)=1540\)

\(207x+621=1540\)

\(207x=1540-621=919\Rightarrow x=\frac{919}{207}\)

a) x=305

b) x=1992

\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)

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

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

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

\(\Rightarrow x+3=308\)

\(\Rightarrow x=305\)

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

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

\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)

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

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

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

\(\Rightarrow x+3=308\)

\(\Leftrightarrow x=308-3\)

\(\Leftrightarrow x=305\)

Vậy \(x=305\)

Tìm x biết :

 a)  \(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)(1)

Gọi B = \(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\left(x+3\right)}\)

     \(\Rightarrow3A=3\left(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\left(x+3\right)}\right)\)  

     \(\Rightarrow3A=\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{x\left(x+3\right)}\)

     \(\Rightarrow3A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\)

     \(\Rightarrow3A=\frac{1}{5}-\frac{1}{x+3}\)

      \(\Rightarrow A=\left(\frac{1}{5}-\frac{1}{x+3}\right):3\)(2)

Thay (2) vào (1) , ta có : \(\left(\frac{1}{5}-\frac{1}{x+3}\right):3=\frac{101}{1540}\)

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

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

                                 \(\Rightarrow x+3=308\)

                                 \(\Rightarrow x=305\)

                                  Vậy x=305

Tương tự như câu b) nha 

Chúc bạn học tốt!! nhớ mik nha

a) Đặt \(A=\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+.....+\frac{1}{\left(x-2\right)x}+\frac{1}{x\left(x+2\right)}\)

=> \(3A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+.....+\frac{3}{\left(x-2\right)x}+\frac{3}{x\left(x+2\right)}\)

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

=> 3A = \(\frac{1}{5}-\frac{1}{x+2}\)

=> A = \(\frac{1}{15}-\frac{1}{3x+6}\)

Mà : A = \(\frac{101}{1540}\)

=> \(\frac{1}{15}-\frac{1}{3x+6}=\frac{101}{1540}\)

=> \(\frac{1}{3x+6}=\frac{1}{15}-\frac{101}{1540}=\frac{1}{924}\)

=> 3x + 6 = 924

=> 3(x + 2) = 924

=> x + 2 = 308

=> x = 306

a) Ta có: \({{1} \over x(x+2)}= {{1} \over 3}({{1} \over x}-{{1} \over x+2})\)  \(\Rightarrow\) \({{1} \over 3}({{1} \over 5}-{{1} \over 8}+{{1} \over 8}-...+{{1} \over x}-{{1} \over x+2})={{101} \over 1540} \)\(\Leftrightarrow\) \({{1} \over 3}({{1} \over 5}-{{1} \over x+2})={{101} \over 1540}\)\(\Leftrightarrow\)x+2 = 308 \(\Leftrightarrow\) x=306 Lúc sau lm hơi tắt mọi người thông cảm

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

\(\Leftrightarrow\)\(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x-3}-\frac{1}{x}+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

\(\Leftrightarrow\)\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}=\frac{303}{1540}\)

\(\Leftrightarrow\)\(\frac{1}{x+3}\)= \(\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)

\(\Leftrightarrow\)\(x+3=308\)

\(\Leftrightarrow\)\(x=308-3=305\)

Vậy   x =305

Mình không viết lại đề bài nha

a) \(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

\(\Rightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

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

\(\Rightarrow\frac{1}{x+3}=\frac{1}{308}\Rightarrow x=305\)

Tìm x,y thuộc Z biết:

a, \(2^{x+y}=2^x+2^y\)

b, \(x+y=x.y=x:y\left(y\ne0\right)\)

Làm nhanh giùm mình nhé!!!!!

1

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

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

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

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

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

\(\Rightarrow x+3=308\)

\(\Rightarrow x=305\)

Vậy x = 305

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

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

\(\dfrac{1}{3}\)(\(\dfrac{1}{5}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{11}\)+...+\(\dfrac{1}{x}\)-\(\dfrac{1}{x+3}\))=\(\dfrac{101}{1540}\)

\(\dfrac{1}{3}\)(\(\dfrac{1}{5}\)-\(\dfrac{1}{x+3}\))=\(\dfrac{101}{1540}\)

\(\dfrac{1}{5}\)-\(\dfrac{1}{x+3}\)=\(\dfrac{101}{1540}\) : \(\dfrac{1}{3}\)

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

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

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

<=>1(x+3)=308.1

<=>1(x+3)=308

<=> x+3=308:1

<=> x+3=308

<=> x=308-3

<=> x=305

b,1+\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+...+\(\dfrac{1}{x\left(x+1\right):2}\)=1\(\dfrac{1991}{1993}\)

\(\dfrac{2}{2}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+3\right)}=\dfrac{3984}{1993}\)\(2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{3984}{1993}\)

\(2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{3984}{1993}\)

\(2\left(1-\dfrac{1}{x+1}\right)=\dfrac{3984}{1993}\)

\(1-\dfrac{1}{x+1}=\dfrac{3984}{1993}:2\)

\(1-\dfrac{1}{x+1}=\dfrac{1992}{1993}\)

\(\dfrac{1}{x+1}=1-\dfrac{1992}{1993}\)

\(\dfrac{1}{x+1}=\dfrac{1}{1993}\)

<=>1(x+1)=1993.1

<=>1(x+1)=1993

<=> x+1=1993 : 1

<=> x+1=1993

<=> x=1993-1

<=> x=1992

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

\(\Rightarrow3\left(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}\right)=3.\frac{101}{1540}\)

\(\Rightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)

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

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

\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)

\(\Rightarrow x+3=308\)

\(\Rightarrow x=305\)

\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\) (x khác 0; khác -3)

\(\Leftrightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)

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

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

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

=>x+3=308

<=>x=305 (nhận)

Vậy x=305