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

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

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

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

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

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

\(\Rightarrow x+3=308\)

\(\Rightarrow x=305\)

vậy \(x=305\)

Ta có: \(\dfrac{1}{3.3}\left(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{101}{1540}\)

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

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

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

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

\(\Rightarrow x+3=308\Rightarrow x=305\)

Ta có:

\(\dfrac{1}{5\times8}+\dfrac{1}{8\times11}+\dfrac{1}{11\times14}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)

\(\dfrac{1}{3}\left(\dfrac{3}{5\times8}+\dfrac{3}{8\times11}+\dfrac{3}{11\times14}+...+\dfrac{1}{x\left(x+3\right)}\right)=\dfrac{101}{1540}\)

\(\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)\(\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)
\(\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{101}{1540}:\dfrac{1}{3}\)
\(\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\dfrac{1}{x+3}=\dfrac{1}{5}-\dfrac{303}{1540}\)
\(\dfrac{1}{x+3}=\dfrac{1}{308}\)
=> x + 3 = 308
x = 308 - 3
x = 305
Vậy x = 305

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

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

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

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

\(=>\dfrac{x+3-5}{5.\left(x+3\right)}=\dfrac{303}{1540}\)

\(=>\dfrac{x-2}{5x+15}=\dfrac{303}{1540}\)

\(=>1540.\left(x-2\right)=303.\left(5x+15\right)\)

\(=>1540x-3080=1515x+4545\)

\(=>1540x-1515x=3080+4545\)

\(=>25x=7625\)

\(=>x=305\)

Vậy x = 305

tick cho mk nha

co gì chưa hiểu thì hỏi nha

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

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

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

\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)

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

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

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

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

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

=> \(\dfrac{1}{x+3}\)= \(\dfrac{5}{1540}=\dfrac{1}{308}\)

=> x+3 = 308

=> x= 305

Vậy x= 305

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

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

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

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

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

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

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

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

⇔ \(\dfrac{1}{x+3}=\dfrac{1}{308}\)

⇔ \(x+3=308\)

⇔ \(x=305\left(TM\right)\)

Vậy ,...

⇒

⇒ 1/5 - 1/ x+3 = 101/1540 .

⇒ 1/5 - 101/1540 = 1/ x+3 .

⇒ 308/1540 - 101/1540 = 1/ x+3 .

⇒ 1/ x+3 = 207/1540 .

⇒ 1540 = ( x + 3 ).207 .

⇒ 1540 = 207x + 621 .

⇒ 1540 - 621 = 207x .

⇒ 207x = 1119 .

⇒ x = 1119 : 207 .

⇒ Không có giá trị của x ( vì x ∈ Z ) .

bài này dễ thế mà ko làm được

\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x.\left(x+3\right)}=\dfrac{101}{1540}\)\(\dfrac{1}{3}.\left(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{x.\left(x+3\right)}\right)=\dfrac{101}{1540}\)\(\dfrac{1}{3}.\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)\(\dfrac{1}{3}.\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)

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

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

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

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

\(\Rightarrow x+3=308\)

\(x=308-3\)

\(x=305\)

Vậy x = 305

a)

<=> (1/3)[3/(5.8) + 3/(8.11) + ... + 3/[x(x+3)] = 101/1540
<=> (1/3)[(1/5 - 1/8) + (1/8 - 1/11) + ... + 1/x - 1/(x+3)] = 101/1540
<=> (1/3)[1/5 - 1/(x+3)] = 101/1540
<=> 1/5 - 1/(x+3) = 303/1540
<=> 1/(x+3) = 1/5 - 303/1540 = 5/1540 = 1/308
<=> x = 305

b)

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

\(\dfrac{1.3}{5.8}+\dfrac{1.3}{8.11}+\dfrac{1.3}{11.14}+...+\dfrac{1.3}{x.\left(x+3\right)}=\dfrac{101.3}{1540}\)

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

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

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

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

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

308.1 = (x + 3).1

308 = x + 3

x = 308 - 3

x = 305

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

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

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

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

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

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

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

=> \(x+3=308\)

=> \(x=305\)

đề bài đúng đấy bạn,hông sai đâu,có cần mk giải lun cho hông???

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

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

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

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

\(\Rightarrow\dfrac{x+3-5}{5\left(x+3\right)}=\dfrac{303}{1540}\)

\(\Rightarrow\dfrac{x-2}{5\left(x+3\right)}=\dfrac{303}{1540}\)

\(\Rightarrow1540\left(x-2\right)=303.5\left(x+3\right)\)

\(\Rightarrow1540x-3080=1515x+4545\)

\(\Rightarrow25x=7625\)

\(\Rightarrow x=305\)

Vậy x = 305

\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+.....+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{1}{3}\left(\dfrac{3}{5.8}+\dfrac{3}{8.11}+.....+\dfrac{3}{x\left(x+3\right)}\right)=\dfrac{101}{1540}\\ \Rightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+....+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\\ \Rightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\\ \Rightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{101}{1540}:\dfrac{1}{3}\\ \Rightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\\ \Rightarrow\dfrac{1}{x+3}=\dfrac{1}{5}-\dfrac{303}{1540}\\ \Rightarrow\dfrac{1}{x+3}=\dfrac{1}{308}\\ \Rightarrow x+3=308\\ \Rightarrow x+=308-3=5\\ \\ \)

a/ \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+.......+\dfrac{1}{2^{10}}\)

\(\Leftrightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+.......+\dfrac{1}{2^9}\)

\(\Leftrightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+......+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{10}}\right)\)

\(\Leftrightarrow A=1-\dfrac{1}{2^{10}}\)

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

\(\Leftrightarrow3\left(\dfrac{1}{5.8}+\dfrac{1}{8.11}+......+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{101}{1540}.3\)

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

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

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

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

\(\Leftrightarrow x+3=308\)

\(\Leftrightarrow x=305\)

Vậy ..

c/ \(1+\dfrac{1}{3}+\dfrac{1}{6}+........+\dfrac{1}{x\left(x+1\right):2}=1\dfrac{2007}{2009}\)

\(\dfrac{1}{2}\left(\dfrac{1}{3}+\dfrac{1}{6}+.......+\dfrac{1}{x\left(x+1\right):2}\right)=\dfrac{4016}{2009}.\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+......+\dfrac{1}{x\left(x+1\right)}=\dfrac{2008}{2009}\)

\(\Leftrightarrow\dfrac{1}{1.2}+\dfrac{1}{2.3}+......+\dfrac{1}{x\left(x+1\right)}=\dfrac{2008}{2009}\)

\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2008}{2009}\)

\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{2008}{2009}\)

\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2009}\)

\(\Leftrightarrow x+1=2009\)

\(\Leftrightarrow x=2008\)

Vậy ..

bài 1:

A=\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)

ta thấy 2A=\(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^9}\)

=>2A-A=\(1-\dfrac{1}{2^{10}}=\dfrac{1023}{1024}\)