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25 tháng 2 2019
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 \(\frac{1}{5}-\frac{1}{x+3}\)= \(\frac{303}{1540}\)=)\(\frac{1}{x+3}=\frac{1}{305}\)=) \(x+3=305\)=) \(x=302\)
TL
3 tháng 9 2015
a) 1/5.8+1/8.11+1/11.14+......+1/x.(x+3)=101/1540
1/3.3.[1/5.8+1/8,11+1/11.14+......+1/x.(x+3)=101/1540
1/3.[3/5.8+3/8.11+3/11.14+........+3/x.(x+3)]=101/1540
1/3.[1/5-1/8+1/8-1/11+1/11-1/14+....+1/x-1/x+3=101/1540
1/3.[1/5-1/x+3]=101/1540
1/5-1/x+3=101/1540.3
1/5-1/x+3=303/1540
1/x+3=1/3-303/1540=1/308
=>x+3=308 =>x=305
Vậy x=305
25 tháng 3 2016
1/3.3(1/5.8+1/8.11+1/11.14+.....1/x(x+1)_101/1540
1/3.(1/5-1/8+1/8-1/11+1/11-1/14+....1/x+1/x+3)=101/1540
1/3.(1/5-1/x+3)=101/1540
1/5-1/x+3=101/1540/1/3=303/1540
1/x+3=1/5-303/1540=1/308
x+3+308
x=305
7 tháng 8 2017
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
Q
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)
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
26 tháng 12 2017
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
26 tháng 12 2017
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
21 tháng 2 2017
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
4 tháng 7 2018
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