Tìm x :
a) \(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
b) \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
K
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R
15 tháng 8 2019
a)1/5.8+1/8.11+1/11.14+...+1/x(x+3)=101/1540
<=>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
<=>1/5-1/x+3=303/1540<=>1/x+3=1/308
<=>x+3=308<=>x=305
Nguồn CHTT, hihi !
HH
22 tháng 3 2016
\(A=\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x.\left(x+3\right)}\Leftrightarrow A=3.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)
\)
\(\Leftrightarrow A=3.\left(\frac{1}{5}-\frac{1}{x+3}\right)\)
Không có gtri A=? ak bạn??
S
18 tháng 3 2017
a, \(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{308}\)
=> x + 3 = 308
x = 308 - 3
x = 305
b, \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=1\frac{1991}{1993}\)
\(\Rightarrow\frac{1}{2}\left(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}\right)=\frac{1}{2}.\frac{3984}{1993}\)
\(\Rightarrow\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{1992}{1993}\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1992}{1993}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1992}{1993}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{1992}{1993}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{1992}{1993}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{1993}\)
=> x + 1 = 1993
x = 1993 - 1
x = 1992
N
18 tháng 3 2017
a ,\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(3.\left(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}\right)=\frac{101}{1540}.3\)
\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{308}\)
\(\Rightarrow x+3=308\)
\(x=308-3\)
\(x=305\)
25 tháng 3 2019
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{1}{3.7}+\frac{1}{4.7}+\frac{1}{4.9}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{2.3.7}+\frac{2}{2.4.7}+\frac{2}{2.4.9}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{6}-\frac{2}{7}+\frac{2}{7}-\frac{2}{8}+....+\frac{2}{x}-\frac{2}{x+1}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{6}-\frac{2}{x+1}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{x+1}=\frac{2}{6}-\frac{2}{9}\)
\(\Rightarrow\frac{2}{x+1}=\frac{1}{3}-\frac{2}{9}\)
\(\Rightarrow\frac{2}{x+1}=\frac{3}{9}-\frac{2}{9}\)
\(\Rightarrow\frac{2}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{2}{x+1}=\frac{2}{18}\)
\(\Rightarrow x+1=18\)
\(\Rightarrow x=17\)
25 tháng 3 2019
câu a khó quá.Để nghĩ.
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{21\cdot2}+\frac{2}{28\cdot2}+\frac{2}{36\cdot2}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow2\left(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)
\(\Rightarrow\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+....+\frac{1}{x\left(x-1\right)}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{x-5}{6x+6}=\frac{1}{9}\)
\(\Rightarrow9\left(x-5\right)=6x+6\)
\(\Rightarrow9x-45=6x+6\)
\(\Rightarrow9x-6x=51\)
\(\Rightarrow3x=51\)
Tới đây bí:v
TN
6 tháng 7 2016
a)\(\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x\left(x+2\right)}=\frac{2}{9}\)
\(2\left(\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{x\left(x+2\right)}\right)=\frac{2}{9}\)
\(2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{2}{9}\)
\(\frac{1}{6}-\frac{1}{x+2}=\frac{2}{9}:2\)
\(\frac{1}{x+2}=\frac{1}{6}-\frac{1}{9}\)
\(\frac{1}{x+2}=\frac{1}{18}\)
=>x+2=18
=>x=16
b tương tự nhân nó với 1/2
D
9 tháng 3 2016
\(\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{65.68}\right)x=\frac{19}{68}+\frac{7}{34}\)
\(\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...-\frac{1}{68}\right)x=\frac{33}{68}\)
\(\left(\frac{1}{2}-\frac{1}{68}\right)x=\frac{33}{68}\)
\(\frac{33}{68}x=\frac{33}{68}\)
\(x=\frac{33}{68}:\frac{33}{68}=1\)
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 \(\frac{1}{5}-\frac{1}{x+3}\)= \(\frac{303}{1540}\)=)\(\frac{1}{x+3}=\frac{1}{305}\)=) \(x+3=305\)=) \(x=302\)
a)\(\frac{1}{5.8}+\frac{1}{8.11}+........+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{x}-\frac{1}{x+3}\right)\)=\(\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)\)
=\(\frac{101}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}\)=\(\frac{101}{1540}:\frac{1}{3}\)=\(\frac{303}{1540}\)
\(\frac{1}{x+3}\)=\(\frac{1}{5}-\frac{303}{1540}\)=\(\frac{1}{308}\)
\(\Rightarrow\)x+3=308
\(\Rightarrow\)x=308-3=305
b)Mk chưa nghĩ ra
b) \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{1}{2}.\frac{2}{9}\)
\(\Rightarrow\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{x+1-6}{6\left(x+1\right)}=\frac{1}{9}\)
\(\Rightarrow\frac{x-5}{6x+6}=\frac{1}{9}\)
\(\Rightarrow9x-45=6x+6\)
\(\Rightarrow3x=51\)
\(\Rightarrow x=17\)
Vậy x = 17