TÍNH
1/5.8+1/8.11+1/11.14+........+1/x(x+3)=101/1540
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\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
=> \(\frac{1}{x+3}=\frac{1}{308}\)
=> x + 3 = 308
=> x = 308 - 3
=> x = 305
\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)
\(\Rightarrow\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{303}{1540}\)
\(\Rightarrow\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\Rightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\Rightarrow\dfrac{1}{x+3}=\dfrac{1}{5}-\dfrac{303}{1540}\)
\(\Rightarrow\dfrac{1}{x+3}=\dfrac{1}{308}\)
\(\Rightarrow x+3=308\)
\(\Rightarrow x=305\)
vậy \(x=305\)
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\Rightarrow3\left(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}\right)=3.\frac{101}{1540}\)
\(\Rightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
\(\Rightarrow x+3=308\)
\(\Rightarrow x=305\)
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\) (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
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
\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\) ( 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 ,...
\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x\left(x+1\right)}\right)=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+1}\right)=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+1}=\frac{303}{1540}\Rightarrow\frac{1}{x+1}=\frac{1}{308}\)
=> x + 1 = 380 => x = 308 - 1 => x = 307
Vậy x = 307
=1/3(3/5.8+3/8.11+............+1/x(x+3)=101/1540
=.1/3(1/5.8+1/8.11+......1/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=101/1540 chia 1/3 =303/1540
=>1/x+3= 1/308
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