5/1.6 +5/6.11 + ...... + 5/101.106
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, bạn tự làm
b, \(B=\dfrac{5^2}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{101}-\dfrac{1}{106}\right)\)
\(=5\left(1-\dfrac{1}{106}\right)=\dfrac{5.105}{106}=\dfrac{525}{106}\)
c, đk : \(x\ne\dfrac{2}{3}\)
Ta có : \(\left|x-1\right|=2\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)(tm)
Với x = 3 suy ra \(C=\dfrac{2.9+9-1}{3.3-2}=\dfrac{26}{7}\)
Với x = -1 suy ra \(C=\dfrac{2-3-1}{-3-2}=\dfrac{-2}{-5}=\dfrac{2}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=5\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{101\cdot106}\right)\\ =5\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{101}-\dfrac{1}{106}\right)\\ =5\left(1-\dfrac{1}{106}\right)=5\cdot\dfrac{105}{106}=\dfrac{525}{106}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=1/5(5/1*6+5/6*11+...+5/101*106)
=1/5(1-1/6+1/6-1/11+...+1/101-1/106)
=1/5(1-1/106)
=1/5*105/106
=21/106
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\dfrac{1}{1.6}+\dfrac{1}{6.11}+\dfrac{1}{11.16}+...+\dfrac{1}{101.106}\)
\(B=\dfrac{1}{5}.\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{101}-\dfrac{1}{106}\right)\)
\(B=\dfrac{1}{5}.\left(1-\dfrac{1}{106}\right)\)
\(B=\dfrac{1}{5}.\dfrac{105}{106}\)
\(B=\dfrac{21}{106}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tính S:
S=5.(\(\dfrac{5}{1.6}\)+\(\dfrac{5}{6.11}\)+...+\(\dfrac{5}{101.106}\))
S=5.(1-\(\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{101}-\dfrac{1}{106}\))
S=5.(1-\(\dfrac{1}{106}\))
S=5.\(\dfrac{105}{106}\)
S=\(\dfrac{525}{106}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
S = 2(5/1.6 + 5/6.11 +.......+ 5/101.106)
S = 2( 1 - 1/6 + 1/6 - 1/11 +.....+ 1/101 - 1/106)
S = 2( 1 - 1/106)
S = 2 . 105/106
S = 105/53
k mk đi,mk mới bị trừ điểm!
1/2.S=5/(1.6)+5/(6.11)+...+5/(101.106)
1/2.S=1/1-1/6+1/6-1/11+...+1/101-1/106
1/2.S=1/1-1/106
1/2.S=105/106
S=105/53
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=2\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{101.106}\right)\)
\(S=2\left(1-\frac{1}{106}\right)\)
\(S=\frac{210}{106}=\frac{105}{53}\)
\(S=2.\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{101.106}\right)\)
\(S=2.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-...-\frac{1}{101}+\frac{1}{101}-\frac{1}{106}\right)\)
\(S=2.\left[1+\left(\frac{-1}{6}+\frac{1}{6}\right)+\left(\frac{-1}{11}\frac{1}{11}\right)+...+\left(\frac{-1}{101}+\frac{1}{101}\right)-\frac{1}{106}\right]\)
\(S=2.\left[1+0+0+...+0-\frac{1}{106}\right]\)
\(S=2.\left[1-\frac{1}{106}\right]\)
\(S=2.\frac{105}{106}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2 phần dưới không liên quan gì đến tính chất trên
a) \(A=\frac{5-2}{2.5}+\frac{8-5}{5.8}+...+\frac{20-17}{17.20}\)
\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)
\(A=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)
b) \(B=5\left(\frac{6-1}{1.6}+\frac{11-6}{6.11}+...+\frac{106-101}{101.106}\right)\)
\(B=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{101}-\frac{1}{106}\right)\)
\(B=5.\left(1-\frac{1}{106}\right)=\frac{525}{106}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`A=-5/(1.6)-5/(6.11)-5/(11.16)-...-5/(2006.2011)`
`-A=5/(1.6)+5/(6.11)+5/(11.16)+...+5/(2006.2011)`
`-A=1-1/6+1/6-1/11+1/11-1/16+.....+1/2006-1/2011`
`-A=1-1/2011=2010/2011`
`A=-2010/2011`
= 1 - 1/6 + 1/6 - 1/11 + ... + 1/101 - 1/106
= 1 - 1/106
= 105/106
Ta có:\(\frac{5}{1.6}+\frac{5}{6.11}+........+\frac{5}{101.106}\)
=\(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+..........+\frac{1}{101}-\frac{1}{106}\)
=\(1-\frac{1}{106}\)
=\(\frac{105}{106}\)