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\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{5\cdot6}=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+...+\left(\frac{1}{5}-\frac{1}{6}\right)=1-\frac{1}{6}=\frac{5}{6}.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
đặt A=1/1x2+1/2x3+1/3x4+1/4x5+1/5x6
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{5}-\frac{1}{6}\)
\(=1-\frac{1}{6}\)
\(=\frac{5}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/1x2 + 1/2x3 + 1/3x4 + 1/4x5 + 1/5x6 + 1/6x7
=1/1-1/2+1/2-1/3+...-1/7
=1+(1/2-1/2+1/3-1/3+...+1/6-1/6)-1/7
=1 +0+0+...-1/7
=1-1/7
=6/7
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a.1/7 + 2/7 + 3/7 + 4/7 + 5/7 + 6/7 = ( 1/7+6/7) + ( 2/7+5/7) + (3/7+4/7)
= 1 + 1 + 1
= 3
b. = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 ( loại bỏ các p/s giống nhau)
= 1/1 - 1/6
= 5/6
a. Các phân số bé hơn 1 có mẫu số bằng 7 là: \(\frac{1}{7};\frac{2}{7};\frac{3}{7};\frac{4}{7};\frac{5}{7};\frac{6}{7}\)
Ta có : \(\frac{1}{7}+\frac{2}{7}+\frac{3}{7}+\frac{4}{7}+\frac{5}{7}+\frac{6}{7}\)
\(=\frac{1+2+3+4+5+6}{7}=\frac{21}{7}=3\)
b. \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=\frac{1}{1}-\frac{1}{6}\)
\(=\frac{5}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/1x2 + 1/2x3 + 1/3x4 + 1/4x5 + 1/5x6
Đặt A = 1/1x2 + 1/2x3 + 1/3x4 + 1/4x5 + 1/5x6
A = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6
A = 1/1 - 1/6
A = 5/6
Vậy A = 5/6
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\(B=\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{50\cdot51}\right)+\left(\dfrac{5}{6}+\dfrac{19}{20}+...+\dfrac{2549}{2550}\right)\)
\(B=\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+..+\dfrac{1}{50\cdot51}\right)+\left(1-\dfrac{1}{2\cdot3}\right)+\left(1-\dfrac{1}{3\cdot4}\right)+...+\left(1-\dfrac{1}{50\cdot51}\right)\)
\(B=\left(1+1+...+1\right)+\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{50\cdot51}\right)-\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{50\cdot51}\right)\)
\(B=1\cdot49=49\) (vì có (50 - 2) : 1 + 1 = 49 số hạng 1)
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}\)
= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)
= \(1-\dfrac{1}{6}\)
= \(\dfrac{5}{6}\)
5/6