a) 1/3-1/4
b) 1/4-1/5
c) S= 1/1.2+ 1/2.3+.........+1/N.(N+1)
d) M= 2/3.5+ 2/5.7+....+ 2/97.99
giải thích cho mik luôn ạ
Tính tổng
a. 1/1.2 + 1/2.3 + 1/3.4 + ...+ 1/2015.2016
b. 2/1.3 + 2/3.5 + 2/5.7 + ...+ 2/99.101
c. 5/1.3 + 5/3.5 + 5/5.7 + ...+ 5/99.101
d. 1/2 + 1/6 + 1/12 + 1/20 + ...+ 1/9900
bn nào trả lời nhanh mik tích, cảm ơn ạ
còn cần không bạn, mk làm cho
a) 1/1.2+1/2.3+1/3.4+...+1/99.100
b) 2/3.5+2/5.7+...+2/49.51
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
bài 1 tính tổng S=1+2+3+...+50
bài 2 cho A là tập hợp các số có 3 chữ số chia cho 4 dư 3
a viết A theo 2 cách
b tính tổng các phần tử của A
bai3 a1.2 +2.3+3.4+...+19.20
b 1.2 +2.3+3.4+...+n [n-1]
bai 4 a 1.2.3 +2.3.4+3.4.5+... +98.99.100
b 1.2.3 +2.3.4 + ...+n[n+1][n+2]
bai 5 a1.3+3.5+5.7+..+97.99
b 2.4+4.6+6.8+...+98.100
Bài 1 Số số hạng của dãy là : (50-1):1+1=50(số hạng )
S = (50+1) x 50 : 2 = 1275
A= 1/1.2+1/2.3+1/3.4+...+1/49.50
A=2/1.3+2/3.5+2/5.7+....+2/49.51
A=1/2.4+1/4.6+1/6.8+....+1/18+20
Giúp e ạ,nhanh e tik ( ◜‿◝ )♡( ◜‿◝ )♡
\(A=\) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}\)
\(A=\frac{49}{50}\)
\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.50}\)
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 = \(\frac{1}{1}-\frac{1}{51}=\frac{50}{51}\)
Tính nhanh
1, A = 1/1.2 + 1/2.3 + 1/3.4 + 1/3.4 + ... + 1/49.50
2, B = 2/3.5 + 2/5.7 + 2/7.9 + ... + 2/37.39
3, C = 5^2/1.6 + 5^2/6.11 + ... + 5/26.31
4, D = 1/7 + 1/91 + 1/247 + 1/475 + 1/775 + 1/1147
x + 25 = 64
x = 64 - 25
x = 39
Vậy x = 39
a,1/1.2 + 1/2.3 + 1/ 3.4+....+ 1/2017.2018
b, 2/3.5+ 2/ 5.7 + 2/7.9 +....+ 2/.91.99
Ai giúp mink tích cho!
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 )
\(=\frac{1}{2}\left(\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{97}-\frac{2}{99}\right)\)
\(=\frac{1}{2}\left(\frac{2}{3}-\frac{2}{99}\right)\)
\(=\frac{1}{2}.\frac{64}{99}\)
\(=\frac{32}{99}\)
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
A=1/1.2+1/2.3+1/3.4+....+1/50.51
B=2/3.5+2/5.7+2/7.9+....+2/51.53
C=6/4.7+6/7.10+6/10.13+....+6/73.76
GIÚP MIK VỚI CẢM ƠN MN.
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
B=6/4.7+6/7.10+6/10.13+...+6/73.76
=2.(3/4.7 +3/7.10 +3/10.13 +...+3/73.76 )
=2.(1/4 −1/7 +1/7 −1/10 +1/10 −1/13 +...+1/73 −1/76 )
=2.(1/4-1/76)=2.9/38=9/19
Tính tổng
A=1/1.2+1/2.3+1/3.4+1/5.6
B=2/1.3+2/3.5+2/5.7+...+2/99.101
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=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)=
\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...\dfrac{1}{99}+\dfrac{1}{101}\)
=\(1-\dfrac{1}{101}\)
=\(\dfrac{100}{101}\)
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\(\left(1-\dfrac{1}{101}\right)\)
= \(\dfrac{200}{101}\)
C= 1/1.2 + 1/2.3 + 1/3.4 +........+ 1/999.1000 F= 1/1.3 + 1/3.5 + 1/5.7 +......+ 1/99.1000
Nêu cách làm nha
Cách làm :
Áp dụng công thức : \(\dfrac{n}{a\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\)
\(C=\dfrac{1}{1.2}+\dfrac{1}{2.3}+..........+\dfrac{1}{999.1000}\)
\(\Leftrightarrow C=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..........+\dfrac{1}{999}-\dfrac{1}{1000}\)
\(\Leftrightarrow C=1-\dfrac{1}{1000}\)
\(\Leftrightarrow C=\dfrac{999}{1000}\)
\(F=\dfrac{1}{1.3}+\dfrac{1}{3.5}+.........+\dfrac{1}{99.101}\)
\(\Leftrightarrow2F=\dfrac{2}{1.3}+\dfrac{2}{3.5}+............+\dfrac{2}{99.101}\)
\(\Leftrightarrow2F=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+........+\dfrac{1}{99}-\dfrac{1}{101}\)
\(\Leftrightarrow2F=1-\dfrac{1}{101}\)
\(\Leftrightarrow2F=\dfrac{100}{101}\)
\(\Leftrightarrow F=\dfrac{50}{101}\)
Giải:
\(C=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{999.1000}\)
\(\Leftrightarrow C=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{999}-\dfrac{1}{1000}\)
\(\Leftrightarrow C=\dfrac{1}{1}-\dfrac{1}{1000}\)
\(\Leftrightarrow C=\dfrac{999}{1000}\)
Sửa đề:
\(F=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{999.1001}\)
\(\Leftrightarrow F=\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{999}-\dfrac{1}{1001}\right)\)
\(\Leftrightarrow F=\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{1001}\right)\)
\(\Leftrightarrow F=\dfrac{1}{2}.\dfrac{1000}{1001}\)
\(\Leftrightarrow F=\dfrac{500}{1001}\)
Chúc bạn học tốt!
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{999.1000}\)
=\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{999}-\dfrac{1}{1000}\)(Áp dụng t.c\(\dfrac{1}{a\left(a+1\right)}=\dfrac{1}{a}-\dfrac{1}{a+1}\))
=\(\dfrac{1}{1}-\dfrac{1}{1000}=\dfrac{999}{1000}\)
Vậy...
\(F=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{997.999}\)
=>\(2F=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{997.999}\)
=>\(2F=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{997}-\dfrac{1}{999}\)(áp dụng tính chất \(\dfrac{2}{a\left(a+2\right)}=\dfrac{1}{a}-\dfrac{1}{a+2}\))
=>\(2F=\dfrac{1}{1}-\dfrac{1}{999}=\dfrac{998}{999}\)
=>\(F=\dfrac{499}{999}\)
Vậy...