NP
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K
Khách
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28 tháng 2 2015
1 + 1/3 + 1/6 + 1/10 + .......... + 1/x.(x+1):2 =1 + 1991/1993
1/2.(1 + 1/3 + 1/6 + 1/10+........+ 1/x.(x+1):2=3984/3986
1/2 + 1/6 +1/12 + .......... +1/x.(x+1)=3984/3986
1/1.2 + 1/2.3 + 1/3.4 +..........+.1/x.(x+1)=3984/3986
2-1/1.2 + 3-2/2.3 + 4-3/3.4 +..........+ x + 1 - x/x.(x+1)
1-1/2+1/2-1/3+1/3-1/4+..........+1/x -1/x+1 =3984/3986
1-1/x+1=3984/3986
1/x+1=1-3984/3986
1/x+1=2/3986=1/1993
x+1=1993
x =1993-1
x =1992
TN
25 tháng 11 2016
a)\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{308}\)
\(\Leftrightarrow x+3=308\Leftrightarrow x=305\)
b ko hiểu đề
NT
0
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
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{\frac{x\left(x+1\right)}{2}}=1\frac{1993}{1991}\)
\(\Leftrightarrow\left(1\cdot\frac{1}{2}\right)+\left(\frac{1}{3}\cdot\frac{1}{2}\right)+\left(\frac{1}{6}\cdot\frac{1}{2}\right)+....+\left(\frac{1}{\frac{x\left(x+1\right)}{2}}\cdot\frac{1}{2}\right)=1\frac{1993}{1991}\div2\)
\(\Leftrightarrow\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{x\left(x+1\right)}=\frac{1992}{1991}\)
\(\Leftrightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{1992}{1991}\)
\(\Leftrightarrow\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1992}{1991}\)
\(\Leftrightarrow\frac{1}{x+1}=1-\frac{1992}{1991}\)
\(\Leftrightarrow\frac{1}{x+1}=-\frac{1}{1991}\)
\(\Leftrightarrow x=-1992\)