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LH
8 tháng 3 2020
\(A=\) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}\)
\(A=\frac{49}{50}\)
LH
8 tháng 3 2020
\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.50}\)
A= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
A = \(\frac{1}{1}-\frac{1}{51}=\frac{50}{51}\)
SK
2
LN
26 tháng 4 2017
A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{5.6}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\)
=1\(-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\)
=\(\dfrac{47}{60}\)
B=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)=
\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...\dfrac{1}{99}+\dfrac{1}{101}\)
=\(1-\dfrac{1}{101}\)
=\(\dfrac{100}{101}\)
25 tháng 4 2017
A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{5.6}\)
= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\)
=\(1-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\)
= \(\dfrac{47}{60}\)
B= \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
= \(2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
= 2\(\left(1-\dfrac{1}{101}\right)\)
= \(\dfrac{200}{101}\)
DL
27 tháng 2 2018
a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2017\cdot2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
b) \(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{97\cdot99}\)( sửa 91.99 thành 97.99 mới đúng nha )
\(=\frac{1}{2}\left(\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{97}-\frac{2}{99}\right)\)
\(=\frac{1}{2}\left(\frac{2}{3}-\frac{2}{99}\right)\)
\(=\frac{1}{2}.\frac{64}{99}\)
\(=\frac{32}{99}\)
AF
27 tháng 2 2018
a) 1/1.2 + 1/2.3 + 1/3.4 +...+1/2017.2018
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ....+1/2017 - 1/2018
= 1 - 1/2018
= 2017/2018
1 tháng 3 2017
A=1/1-1/2+1/2-1/3+1/3-1/4+....+1/49-1/50
A=1/1-1/50
A=49/50
Vay A=49/50
B=1/3-1/5+1/5-1/7....+1/37-1/39
B=1/3-1/39
b=36/117
B=4/13
M
16 tháng 1 2020
Bài giải
\(B=1\cdot2^2+2\cdot3^2+3\cdot4^2+...+99\cdot100^2\)
\(B=1\cdot2\cdot\left(3-1\right)+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-1\right)+...+99\cdot100\cdot\left(101-1\right)\)
\(B=1\cdot2\cdot3-1\cdot2+2\cdot3\cdot4-2\cdot3+...+99\cdot100\cdot101-99\cdot100\)
\(B=\left(1\cdot2\cdot3+2\cdot3\cdot4+...+99\cdot100\cdot101\right)-\left(1\cdot2+2\cdot3+...+99\cdot100\right)\)
Đặt \(C=1\cdot2\cdot3+2\cdot3\cdot4+...+99\cdot100\cdot101\)
\(4C=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot\left(5-1\right)+...+99\cdot100\cdot101\cdot\left(102-98\right)\)
\(4C=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4+...+99\cdot100\cdot101\cdot102-98\cdot99\cdot100\cdot101\)
\(4C=99\cdot100\cdot101\cdot102\)
\(4C=101989800\)
\(C=101989800\text{ : }4\)
\(C=25497450\)
T
5 tháng 5 2020
A = 1 /1.2 + 1/ 2.3 + 1 /3.4 + . . . + 1/ 49.50 + 1/ 50.51
A = 2 − 1/ 1.2 + 3 − 2 /2.3 + 4 − 3 /3.4 + . . . + 50 − 49 /49.50 + 51 − 50/ 50.51
A = 1 − 1/ 2 + 1/ 2 − 1 /3 + 1 /3 − 1/ 4 + . . . + 1 /50 − 1 /51
A=1-1/51
A=50/51
4 tháng 2 2017
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
b) \(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=2.\left(1-\frac{1}{99}\right)\)
\(=2.\frac{98}{99}\)
\(=\frac{196}{99}=1\frac{97}{99}\)
a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
b) \(\frac{2}{3\cdot5}+\frac{3}{5\cdot7}+...+\frac{2}{49\cdot51}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(=\frac{1}{3}-\frac{1}{51}\)
\(=\frac{16}{51}\)
a) 1/1.2+1/2.3+1/3.4+...+1/99.100
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 -1/4 + ... + 1/99 - 1/100
= 1/1 - 1/100
= 99/100
b) 2/3.5+2/5.7+...+2/49.51
= 2 . ( 1/3.5 + 1/5.7 + ... + 1/49.51 )
= 2 . ( 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/49 - 1/50 )
= 2 . ( 1/3 - 1/50 )
= 2 . 47/150
= 47/75