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.
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 = \(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)
= \(3\left(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{73.76}\right)\)
= \(3.\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\right)\)
= \(\frac{3}{3}\left(\frac{1}{4}-\frac{1}{76}\right)\)
= \(\frac{9}{38}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
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
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 \(D=\frac{10}{39}\)
a)6/1.4+6/4.7+6/7.10+....+6/97.100
b 4/1.3+16/3.5+36/5.7+.....+9604/97.99
c (1/1.2+1/3.4+1/5.6+...........+1/19.20)-(1/11+1/12+.....+1/20)
a)Đặt \(A=\dfrac{6}{1.4}+\dfrac{6}{4.7}+\dfrac{6}{7.10}+...+\dfrac{6}{97.100}\)
\(3a=3-\dfrac{3}{4}+\dfrac{3}{4}-\dfrac{3}{7}+\dfrac{3}{7}-\dfrac{3}{10}+...+\dfrac{3}{97}-\dfrac{3}{100}\)
\(=3-\dfrac{3}{100}\)
\(=\dfrac{297}{100}\)
b)Đặt \(B=\dfrac{4}{1.3}+\dfrac{16}{3.5}+\dfrac{36}{5.7}+...+\dfrac{9604}{97.99}\)
\(=2b=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\)
\(2b=2-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{2}{7}+...+\dfrac{2}{97}-\dfrac{2}{99}\)
\(2b=2-\dfrac{2}{99}=\dfrac{198}{99}-\dfrac{2}{99}=\dfrac{196}{99}\)
c) Tương tự! Bạn tự làm nhé!
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
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
a) \(A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)
\(A=\frac{1}{3}-\frac{1}{39}\)
\(A=\frac{13}{39}-\frac{1}{39}=\frac{12}{39}\)
b) \(B=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)
\(B=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\)
\(B=\frac{1}{4}-\frac{1}{76}\)
\(B=\frac{19}{76}-\frac{1}{76}=\frac{18}{76}=\frac{9}{38}\)
a, \(A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)
\(A=\frac{1}{3}-\frac{1}{39}\)
\(A=\frac{4}{13}\)
b, \(B=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)
\(B=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\)
\(B=\frac{1}{4}-\frac{1}{76}\)
\(B=\frac{9}{38}\)
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: \(B=\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{73}-\dfrac{1}{76}\)
\(=\dfrac{1}{4}-\dfrac{1}{76}=\dfrac{18}{76}=\dfrac{9}{38}\)
A=1/1.2+1/2.3+1/3.4+1/4.5+...+1/98.99+1/99.100
B=2/1.3+2/3.5+2/5.7+2/7.9+...+2/97.99+2/99.101
C=1/2+1/4+1/8+1/16+...+1/1024+1/2048
D=1/2+1/6+1/18+1/54+1/4374+1/13122 ( . (DẤU CHẤM LÀ ''NHÂN'')
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{!}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
\(C=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....+\frac{1}{1024}+\frac{1}{2048}\)
\(\Rightarrow\)\(2C=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{512}+\frac{1}{1024}\)
\(\Rightarrow\)\(2C-C=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2048}\right)\)
\(\Leftrightarrow\)\(C=1-\frac{1}{2048}=\frac{2047}{2048}\)
Câu A bạn quên 1/4.5 kìa , với câu D đâu >>>
Đề bài :
a) dãy các phân số trên có phải theo quy luật ko ?
b) tính tổng các phân số của dãy trên
1)\(\dfrac{1}{1.2};\dfrac{1}{2.3};\dfrac{1}{3.4};\dfrac{1}{4.5};\dfrac{1}{49.50}\)
2) \(\dfrac{2}{3.5};\dfrac{2}{5.7};\dfrac{2}{7.9};...;\dfrac{2}{37.39}\)
3) \(\dfrac{3}{4.7};\dfrac{3}{7.10};\dfrac{3}{10.13};...;\dfrac{3}{73.76}\)
MIK CẦN GẤP .MN GIÚP MIK MHA !!!
TRÌNH BÀY RA NHA !!! THANKS YOU SO MUCH
Đề bài :
a) dãy các phân số trên có phải theo quy luật ko ?
b) tính tổng các phân số của dãy trên
1) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=1-\dfrac{1}{50}\)
\(=\dfrac{49}{50}\)
2) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{37.39}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{37}-\dfrac{1}{39}\)
\(=\dfrac{1}{3}-\dfrac{1}{39}\)
\(=\dfrac{13}{39}-\dfrac{1}{39}=\dfrac{12}{39}=\dfrac{4}{13}\)
3) \(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{73.76}\)
\(=\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{73}-\dfrac{1}{76}\)
\(=\dfrac{1}{4}-\dfrac{1}{76}\)
\(=\dfrac{19}{76}-\dfrac{1}{76}=\dfrac{18}{76}=\dfrac{9}{38}\)
1)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\\ =1-\dfrac{1}{50}\\ =\dfrac{49}{50}\)
2)
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{37.39}\\ =\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{39}\\ =\dfrac{1}{3}-\dfrac{1}{39}\\ =\dfrac{13}{39}-\dfrac{1}{39}\\ =\dfrac{12}{39}=\dfrac{4}{13}\)
3) \(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{73.76}\\ =\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{73}-\dfrac{1}{79}\\ =\dfrac{1}{4}-\dfrac{1}{79}\\ =\dfrac{75}{316}\)
phần a thêm một câu hỏi nữa là : nếu có quy luật là gì ? .
MONG MN GIÚP ĐỠ Ạ !!!