bài 1 tính
a, A= 2/3.5 + 2/5.7 + 2/7.9 + ......+ 2/37.39
b, B= 3/4.7 + 3/7.10 + 3/10.13 +.......+3/73.76
bài 3 tính tổng
a, A=2/3.5+2/5.7+2/7.9+.....+2/201.203
b, B=3/4.7+3/7.10+3/10.13+.....+3/73.76
mik đang gấp lắm
a: \(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{201}-\dfrac{1}{203}=\dfrac{1}{3}-\dfrac{1}{203}=\dfrac{200}{609}\)
b: \(B=\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{73}-\dfrac{1}{76}\)
\(=\dfrac{1}{4}-\dfrac{1}{76}=\dfrac{18}{76}=\dfrac{9}{38}\)
tính: A=1 phần 1.2 +1 phần 2.3+ 1 phần 3.4+...1 phần 49.50
B=2 phần 3.5+ 2 phần 5.7+ 2 phần 7.9 +...+ 2 phần 37.39
C= 3 phần 4.7 + 3 phần 7.1 + 3 phần 10.13 + ... + 3 phần 73.76
A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
= \(1-\frac{1}{50}=\frac{49}{50}\)
B = \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)
= \(2\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{37.39}\right)\)
= \(2.\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\right)\)
= \(\frac{2}{2}\left(\frac{1}{3}-\frac{1}{39}\right)\)
= \(\frac{4}{13}\)
C = \(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)
= \(3\left(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{73.76}\right)\)
= \(3.\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\right)\)
= \(\frac{3}{3}\left(\frac{1}{4}-\frac{1}{76}\right)\)
= \(\frac{9}{38}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
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 :
a) A = \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+....+\dfrac{2}{37.39}\)
b) B = \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+.....+\dfrac{3}{73.76}\)
a, \(A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{37.39}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{39}\)
\(=\dfrac{1}{3}-\dfrac{1}{39}\)
\(=\dfrac{12}{39}\)
Vậy \(A=\dfrac{12}{39}\)
b,\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{73.76}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{73}-\dfrac{1}{76}\)
\(=1-\dfrac{1}{76}\)
\(=\dfrac{75}{76}\)
Vậy \(B=\dfrac{75}{76}\)
a) Ta có :
\(A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+....................+\dfrac{2}{37.39}\)
\(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...................+\dfrac{1}{37}-\dfrac{1}{39}\)
\(A=\dfrac{1}{3}-\dfrac{1}{39}=\dfrac{4}{13}\)
b) Ta có :
\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+..................+\dfrac{3}{73.76}\)
\(B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+..................+\dfrac{1}{73}-\dfrac{1}{76}\)
\(B=1-\dfrac{1}{76}=\dfrac{75}{76}\)
~ Học tốt ~
tính nhanh : A = \(\frac{2}{2.5}+\frac{2}{5.7}+\frac{2}{7.9}+.......+\frac{2}{37.39}\)
B = \(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+......+\frac{1}{73.76}\)
http://olm.vn/hoi-dap/question/772291.html
sau 3 phút có kết quả tùy bạn
Bài 1
a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\)\(\frac{1}{99.100}\) b)\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2015.2017}\)
c)\(\left(\frac{3}{2}+\frac{7}{6}+\frac{13}{12}+...+\frac{9901}{9900}\right)\)\(-\frac{99}{100}\) d)\(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)
a, 1/1.2+1/1.3+...+1/99.100
= 1-1/2+1/2-1/3+1/3+...+1/99-1/100
=1-1/100
=99/100
A = 99/100
B = 2014/6051
C = 99
D = 9/38
Tính tổng:
A=\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
B=\(\frac{2}{4.7}+\frac{2}{7.10}+\frac{2}{10.13}+...+\frac{2}{100.103}\)
\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{99\cdot101}\)
\(A=\frac{1}{2}\cdot\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{99\cdot101}\right)\)
\(A=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(A=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{101}\right)=\frac{1}{2}\cdot\frac{97}{303}=\frac{97}{606}\)
\(B=\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+\frac{2}{10\cdot13}+...+\frac{2}{100\cdot103}\)
\(B=\frac{2}{3}\cdot\left(\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+...+\frac{3}{100\cdot103}\right)\)
\(B=\frac{2}{3}\cdot\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(B=\frac{2}{3}\cdot\left(\frac{1}{4}-\frac{1}{103}\right)=\frac{2}{3}\cdot\frac{99}{412}=\frac{33}{206}\)
B= 1 trên 3.4 + 1 trên 4.5 + 1 trên 5. 6 + 1 trên 6.7 +1 trên 7.8
C= 1 trên 4.5 +1 trên 5.6 + 1 trên 6.7 + 1 trên 7.8 +1 trên 8.9
D= 2 3.5 + 2 5.7 + 2 7.9 + 2 9.11 +2 11.13
E = 3 1.4 + 3 4.7 + 3 7.10 + 3 10.13
Ta có: \(B=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{3}-\frac{1}{8}\)
\(=\frac{5}{24}\)
Vậy \(B=\frac{5}{24}\)
Ta có: \(C=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{4}-\frac{1}{9}\)
\(=\frac{5}{36}\)
Vậy \(C=\frac{5}{36}\)
Ta có: \(C=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)
\(=\frac{1}{3}-\frac{1}{13}\)
\(=\frac{10}{39}\)
Vậy \(D=\frac{10}{39}\)
* Tính :
a) A = 1/3 + 1/15 + 1/35 + 1/63 + 1/99
b) B = 7/4 . ( 33/12 + 33/20 + 33/30 + 33/42 + 33/56 )
c) C = 1/1.3 + 1/3.5 + 1/5.7 + ... + 1/2013.2019
d) D = 3^2/1.4 + 3^2/4.7 + 3^2/7.10 + 3^2/10.13 + 3^2/13.16
* Cho abc = 105, bc + b + 1 khác 0. Tính : S = 105/abc + ab + a + b/bc + b + 1 + a/ab + a + 105
A= \(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{35}+\frac{1}{99}=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.6}+...+\frac{2}{9.11}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\)
\(2A=1-\frac{1}{11}=\frac{10}{11}\)
\(A=\frac{10}{11}:2=\frac{5}{11}\)
\(D=\frac{3^2}{1.4}+\frac{3^2}{4.7}+...+\frac{3^2}{13.16}\)
\(D=3.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{13.16}\right)\)
\(D=3.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{16}\right)\)
\(D=3.\left(1-\frac{1}{16}\right)=3.\frac{15}{16}=2\frac{13}{16}\)