Tính :
\(B=\dfrac{\dfrac{1}{1.2001}+\dfrac{1}{2.2002}+\dfrac{1}{3.2003}+...+\dfrac{1}{19.2019}}{\dfrac{1}{1.20}+\dfrac{1}{2.21}+\dfrac{1}{3.22}+....+\dfrac{1}{2000.2019}}\)
Những câu hỏi liên quan
Cho \(A=\dfrac{1}{1.18}+\dfrac{1}{2.19}+\dfrac{1}{3.20}+...+\dfrac{1}{2000.2017}\)
\(B=\dfrac{1}{1.2001}+\dfrac{1}{2.2002}+\dfrac{1}{3.2003}+...+\dfrac{1}{17.2017}\)
Tính A/B
16 tháng 7 2023 lúc 12:19
Bài 1 : Tính tổng sau
a) \(A=\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}\)
b) \(B=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+....+\dfrac{1}{23.24}+\dfrac{1}{24.25}\)
c) \(C=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)
`@` `\text {Ans}`
`\downarrow`
`a)`
\(A=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\)
`=`\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
`=`\(\dfrac{1}{3}-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-...-\dfrac{1}{9}\)
`=`\(\dfrac{1}{3}-\dfrac{1}{9}\)
`=`\(\dfrac{2}{9}\)
Vậy, \(A=\dfrac{2}{9}\)
`b)`
\(B=\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{23\cdot24}+\dfrac{1}{24\cdot25}\)
`=`\(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\)
`=`\(\dfrac{1}{5}-\left(\dfrac{1}{6}-\dfrac{1}{6}\right)-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-...-\dfrac{1}{25}\)
`=`\(\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{4}{25}\)
Vậy, \(B=\dfrac{4}{25}\)
`c)`
\(C=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\)
`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
`=`\(1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-...-\dfrac{1}{100}\)
`=`\(1-\dfrac{1}{100}=\dfrac{99}{100}\)
Vậy, \(C=\dfrac{99}{100}\)
Đúng 4
Bình luận (0)
Tính:
a) A dfrac{1}{1.2} + dfrac{1}{2.3} + dfrac{1}{3.4} +...+ dfrac{1}{998.999} + dfrac{1}{999.1000}
b) B dfrac{1}{1.6} + dfrac{1}{6.11} + dfrac{1}{11.16} +...+ dfrac{1}{495.500}
c) C dfrac{1}{1.2.3} + dfrac{1}{2.3.4} + dfrac{1}{3.4.5} +...+ dfrac{1}{998.999.1000}
(Mong mn giúp ạ)
Đọc tiếp
Tính:
a) A = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) +...+ \(\dfrac{1}{998.999}\) + \(\dfrac{1}{999.1000}\)
b) B = \(\dfrac{1}{1.6}\) + \(\dfrac{1}{6.11}\) + \(\dfrac{1}{11.16}\) +...+ \(\dfrac{1}{495.500}\)
c) C = \(\dfrac{1}{1.2.3}\) + \(\dfrac{1}{2.3.4}\) + \(\dfrac{1}{3.4.5}\) +...+ \(\dfrac{1}{998.999.1000}\)
(Mong mn giúp ạ)
a.
$A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{1000-999}{999.1000}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}$
$=1-\frac{1}{1000}=\frac{999}{1000}$
Đúng 1
Bình luận (0)
b.
$5B=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+....+\frac{5}{495.500}$
$=\frac{6-1}{1.6}+\frac{11-6}{6.11}+\frac{16-11}{11.16}+....+\frac{500-495}{495.500}$
$=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+....+\frac{1}{495}-\frac{1}{500}$
$=1-\frac{1}{500}=\frac{499}{500}$
$\Rightarrow B=\frac{499}{500}: 5= \frac{499}{2500}$
Đúng 1
Bình luận (0)
c.
$2C=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{998.999.100}$
$=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{1000-998}{998.999.1000}$
$=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{998.999}-\frac{1}{999.1000}$
$=\frac{1}{1.2}-\frac{1}{999.1000}=\frac{499499}{999000}$
$\Rightarrow C=\frac{499499}{999000}:2=\frac{499499}{1998000}$
Đúng 0
Bình luận (0)
Cho \(A=\dfrac{1}{1.21}+\dfrac{1}{2.22}+\dfrac{1}{3.23}+...+\dfrac{1}{80.100}\);
\(B=\dfrac{1}{1.81}+\dfrac{1}{2.82}+\dfrac{1}{3.83}+...+\dfrac{1}{20.100}\).
Tính \(\dfrac{A}{B}\).
A=20/1.21+20/2.22+...+20/80.100
=1-1/21+1/2-1/22+...+1/80-1/100
=(1+1/2+...+1/80)-(1/21+1/22+...+1/100)
80B=80/1.81+80/2.82+...+8/20.100
=1-1/81+1/2-1/82+...+1/20-1/100
=(1+1/2+...+1/20)-(1/81+1/82+...+1/100)
=(1+1/2+1/3+...+1/20+1/21+1/22+...+1/80)-(1/21+1/22+...1/80+1/81+1/82+...1/100)
=>20A=80B
=>A=4B
Đúng 3
Bình luận (0)
A=1-\(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}+\dfrac{1}{2019}\)
B=\(\dfrac{1}{1010}+\dfrac{1}{1011}+\dfrac{1}{1012}+...+\dfrac{1}{2019}\)
Tính \(^{\left(A-B\right)^{2019}}\)
Tính nhanh:
a, \(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
b, \(\left(-\dfrac{1}{4}+\dfrac{7}{35}-\dfrac{5}{3}\right)-\left(-\dfrac{15}{12}+\dfrac{6}{11}-\dfrac{48}{49}\right)\)
a: Ta có: \(\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{72}\right)\)
\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}-...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
=0
Đúng 0
Bình luận (1)
Cíuuu
Bài 1: Tính nhanh
1) B= \(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
2) C= \(\dfrac{1}{2}+\dfrac{1}{14}+\dfrac{1}{35}+\dfrac{1}{65}+\dfrac{1}{104}+\dfrac{1}{152}\)
Bài 2: Chứng minh
\(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+\dfrac{1}{299}+\dfrac{1}{300}>\dfrac{2}{3}\)
Tính hợp lý
\(A= (\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\) B= \(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}}{\dfrac{1}{9}+\dfrac{2}{8}+\dfrac{3}{7}+...+\dfrac{8}{2}+\dfrac{9}{1}})\)
Tính nhanh
a, S dfrac{1}{3} + dfrac{1}{15} + dfrac{1}{35} + dfrac{1}{63} + dfrac{1}{99} + dfrac{1}{143}
b, A dfrac{1}{3} + dfrac{1}{6} + dfrac{1}{10} + dfrac{1}{15} + dfrac{1}{21} + dfrac{1}{28}
c, H dfrac{4047991-2010x2009}{4050000-2011x2009}
d, T dfrac{2009x20010+2000}{2011x2010-2020}
e, P dfrac{7589-80,5x69,3}{7485,05-79x69,3}
f, B 5,1 x 42,2 + 1,7 x 448 x 3 - 0,15 x 700
Giúp mình với
Đọc tiếp
Tính nhanh
a, S= \(\dfrac{1}{3}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{35}\) + \(\dfrac{1}{63}\) + \(\dfrac{1}{99}\) + \(\dfrac{1}{143}\)
b, A = \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{21}\) + \(\dfrac{1}{28}\)
c, H =\(\dfrac{4047991-2010x2009}{4050000-2011x2009}\)
d, T = \(\dfrac{2009x20010+2000}{2011x2010-2020}\)
e, P = \(\dfrac{7589-80,5x69,3}{7485,05-79x69,3}\)
f, B = 5,1 x 42,2 + 1,7 x 448 x 3 - 0,15 x 700
Giúp mình với
a=78/35
b=22/12
c=1/1
d=40202090/4040090
e=1,24025667172...
f=871,82
ko biết đúng ko [0_0'] hihi
Đúng 0
Bình luận (0)