K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

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

5 tháng 7 2018

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

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

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

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

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

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

\(\Rightarrow x+3=308\)

\(\Rightarrow x=305\)

vậy \(x=305\)

5 tháng 7 2018

thanksvui

13 tháng 7 2017

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

3 tháng 7 2018

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

3 tháng 7 2018

mình làm xong rồi nhưng dù sao cũng thank

17 tháng 7 2018

Ta có : 1/ 5.8 + 1/ 8.11 + 1/ 11.14 + ... + 1/ x.(x+3) = 101/1540 .

1/5 - 1/8 + 1/8 - 1/11 + ... + 1/x - 1/ x+3 = 101/1540 .

⇒ 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 ) .

23 tháng 7 2018

bài này dễ thế mà ko làm đượckhocroi

25 tháng 3 2017

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

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

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

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

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

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

25 tháng 3 2017

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

24 tháng 1 2022

\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{y\left(y+3\right)}=\dfrac{98}{1545}\)

\(\Leftrightarrow\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{y\left(y+3\right)}=\dfrac{98}{515}\)

\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{y}-\dfrac{1}{y+3}=\dfrac{98}{515}\)

\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{y+3}=\dfrac{98}{515}\)

\(\Leftrightarrow\dfrac{y+3}{5\left(y+3\right)}-\dfrac{5}{5\left(y+3\right)}=\dfrac{98}{515}\)

\(\Leftrightarrow\dfrac{y+3-5}{5\left(y+3\right)}=\dfrac{98}{515}\)

\(\Leftrightarrow\dfrac{y-2}{5\left(y+3\right)}=\dfrac{98}{515}\)

\(\Leftrightarrow515\left(y-2\right)=98.5\left(y+3\right)\)

\(\Leftrightarrow515y-1030=490y+1470\)

\(\Leftrightarrow25y-2500=0\\ \Leftrightarrow25y=2500\\ \Leftrightarrow y=100\)

\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{y+3}\right)=\dfrac{98}{1545}\)

\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{y+3}=\dfrac{98}{515}\)

\(\Leftrightarrow\dfrac{1}{y+3}=\dfrac{1}{103}\)

hay x=100

14 tháng 7 2017

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

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

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

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

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

10 tháng 4 2017

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)

Ôn tập toán 6

10 tháng 4 2017

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

HQ
Hà Quang Minh
Giáo viên
28 tháng 1

Ta có công thức tổng quát: 

\(\dfrac{k}{n\cdot\left(n+k\right)}=\dfrac{1}{n}-\dfrac{1}{n+k}\)

\(a,A=\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{x\left(x+3\right)}\\ =\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{x\left(x+3\right)}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)\\ =\dfrac{1}{3}\cdot\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)\\ =\dfrac{1}{3}\cdot\dfrac{x-2}{5\left(x+3\right)}\\ =\dfrac{x-2}{15\left(x+3\right)}\)

Theo đề bài ta có: 

\(A=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{15\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{303}{308}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{305-2}{305+3}\\ \Rightarrow x=305\)

28 tháng 1

khó nhỉ