Tính nhanh
B = 3/4.8/9.15/16.....2499/2500
K
Khách
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16 tháng 3 2016
B = 3/4.8/9.15/16.....2499/2500
= 1.3/2^2 . 2.4/3^2 . 3.5/4^2 ..... 49.51/50^2
= 1.2.3.....49/1.2.3.....50 . 3.4.5....51/1.2.3.....50
= 1/50 . 51/2
= 51/100
CX
24 tháng 3 2016
Giải
B = 3/4.8/9.15/16.....2499/2500
B=1.3/2^2 . 2.4/3^2 . 3.5/4^2 ..... 49.51/50^2
B=1.2.3.....49/2.3.4.....50 . 3.4.5.....51/2.3.4.....50
B=1/50 . 51/2
B=51/100
NT
2
OY
15 tháng 7 2021
B=\(\dfrac{3}{3.5}.\dfrac{3}{5.7}.....\dfrac{3}{47.49}\)
B=\(\dfrac{3}{2}.\left(\dfrac{2}{3.5}.\dfrac{2}{5.7}.....\dfrac{2}{47.49}\right)\)
B=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{47}-\dfrac{1}{49}\right)\)
B=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{49}\right)\)
B=\(\dfrac{3}{2}.\dfrac{46}{147}\)
B=\(\dfrac{23}{49}\)
15 tháng 7 2021
a) Ta có: \(A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{2499}{2500}\)
\(=\dfrac{1\cdot3}{2^2}\cdot\dfrac{2\cdot4}{3^2}\cdot\dfrac{3\cdot5}{4^2}\cdot...\cdot\dfrac{49\cdot51}{50^2}\)
\(=\dfrac{1}{50}\cdot\dfrac{51}{2}=\dfrac{51}{100}\)
b) Ta có: \(B=\dfrac{3}{3\cdot5}+\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+...+\dfrac{3}{47\cdot49}\)
\(=\dfrac{3}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{47\cdot49}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{49}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{46}{147}=\dfrac{138}{294}=\dfrac{23}{49}\)
AH
Akai Haruma
Giáo viên
14 tháng 9
Lời giải:
\(A=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}....\frac{-998}{999}.\frac{-999}{1000}\\
=\frac{(-1)(-2)(-3)...(-998)(-999)}{2.3.4....1000}\\
=-\frac{1.2.3.4....998.999}{2.3.4...1000}\\
=-\frac{1}{1000}\)
AH
Akai Haruma
Giáo viên
14 tháng 9
Trong $B$ có một thừa số là $1-\frac{7}{7}=0$ nên $B=0$ (do số nào nhân với $0$ cũng sẽ bằng $0$.
----------------------
$C=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{49.51}{50^2}$
$=\frac{1.3.2.4.3.5.....49.51}{2^2.3^2.4^2....50^2}$
$=\frac{(1.2.3...49)(3.4.5...51)}{(2.3.4...50)(2.3.4...50)}$
$=\frac{1.2.3...49}{2.3.4...50}.\frac{3.4.5...51}{2.3.4....50}$
$=\frac{1}{50}.\frac{51}{2}=\frac{51}{100}$
TT
3
XO
11 tháng 6 2020
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{1000}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{999}{1000}=\frac{1.2.3...999}{2.3.4...1000}=\frac{1}{1000}\)
\(B=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{2499}{2500}=\frac{3.8.15...2499}{4.9.16....2500}=\frac{1.3.2.4.3.5....49.51}{2.2.3.3.4.4...50.50}=\frac{\left(1.2.3...49\right).\left(3.4.5...51\right)}{\left(2.3.4...50\right).\left(2.3.4...50\right)}\)
\(\frac{1.51}{50.2}=\frac{51}{100}\)
11 tháng 6 2020
a. \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{999}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{998}{999}\)
\(A=\frac{1\cdot2\cdot3\cdot....\cdot998}{2\cdot3\cdot4\cdot....\cdot999}=\frac{1}{999}\)
Vậy \(A=\frac{1}{999}\)
ND
2
24 tháng 2 2016
17/25
duyet di nhanh
Nguyễn Đức Nhật Minh h mk may man ca nam
20 tháng 3 2018
= ( 1 x 3 / 2 x 2 ) . ( 2 x 4 / 3 x 3 ) . ( 3 x 5 / 4 x 4 ) . . . ( 6 x 8 / 7 x 7 ) ( mk viết thế cho rõ nhìn khi lm bỏ ngoặc cx dc )
= 1 x 3 x 2 x 4 x 3 x 4 x ... x 6 x 8 / 2 x 2 x 3 x 3 x 4 x 4 x ... x 7 x 7
= 1 x 8 / 2 x 7
= 8 / 14
= 4/7
Tk nha !!
NT
20 tháng 3 2018
Ta có :
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}........\frac{48}{49}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.......\frac{6.8}{7.7}\)
\(=\frac{\left(1.2.3.....6\right).\left(3.4.5.....8\right)}{\left(2.3.4......7\right).\left(2.3.4.....7\right)}\)
\(=\frac{1.2.3.....6}{2.3.4.....7}.\frac{3.4.5......8}{2.3.4......7}\)
\(=\frac{1}{7}.4\)
\(=\frac{4}{7}\)
D
22 tháng 4 2015
a)\(\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.....\frac{2499}{2500}\) \(=\frac{2.4}{3^2}.\frac{3.5}{4^2}.\frac{4.6}{5^2}.....\frac{49.51}{50^2}\)\(=\frac{\left(2.3.4.....49\right)\left(4.5.6....51\right)}{\left(3.4.5.....50\right)\left(3.4.5.....50\right)}=\frac{2.51}{50.3}=\frac{1.17}{25}=\frac{17}{25}\)
D
22 tháng 4 2015
b)\(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)......\left(1-\frac{1}{780}\right)\)
\(=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}......\frac{779}{780}\)\(=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.....\frac{1558}{1560}\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.....\frac{38.41}{39.40}\)
\(=\frac{\left(1.2.3.....38\right)\left(4.5.6.....41\right)}{\left(2.3.4.....39\right)\left(3.4.5.....40\right)}=\frac{1.41}{39.3}=\frac{41}{117}\)
19 tháng 12 2017
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{9999}{10000}\)
\(A=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{99.101}{100.100}\)
\(A=\frac{1.2.3...99}{2.3.4...100}.\frac{3.4.5...101}{2.3.4...100}=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)
B=\(\frac{3}{4}\).\(\frac{8}{9}\).....\(\frac{2499}{2500}\)
B=\(\frac{1.3}{2^2}\).\(\frac{2.4}{3^2}\).....\(\frac{49.50}{50^2}\)
B=\(\frac{1.3.2.4....49.50}{2^2.3^2....50^2}\)
B=\(\frac{\left(1.2....49\right).\left(3.4...50\right)}{\left(2.3...50\right).\left(2.3...50\right)}\)
B=\(\frac{1}{50.2}\)=\(\frac{1}{100}\)