Tính A= 1/2 . 1/3 + 1/3 . 1/4 + 1/4 . 1/5 +...+1/99 . 1/100
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NT
0
N
0
DA
23 tháng 3 2020
1. 1 + ( -2) +3 +(-4) + .........+ 19 + (-20)
= -1 + ( -1) +....+(-1)
= -1. 10
= -10
2. 1 – 2 + 3 – 4 + . . . + 99 – 100
= ( -1) + (-1) +....+(-1)
= -1. 50
= -50
3. 2 – 4 + 6 – 8 + . . . + 48 – 50
= (-2) + (-2) +....+ (-2)
= -2. 12 + 26
= -24 + 26
= 2
4. – 1 + 3 – 5 + 7 - . . . . + 97 – 99
= 2 + 2 +......+2
= 2.25
= 50
5. 1 + 2 – 3 – 4 + ... + 97 + 98 – 99 - 100
= (1+2-3-4) +......+ ( 97+98-99 -100)
= -4 . (-4).....(-4)
= -4. 25
= -100
2 tháng 6 2018
a) \(A=98+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\)(có 98 phân số nên ta cộng 1 vào mỗi phân số)
\(A=\left(\frac{1}{2}+1\right)+\left(\frac{1}{3}+1\right)+...+\left(\frac{1}{99}+1\right)\)
\(A=\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}\)
Và \(B=\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}}{\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}}=1\)
b) \(A=2018+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\)(có 2018 phân số nên ta cộng 1 vào mỗi phân số)
\(A=\left(\frac{1}{2}+1\right)+\left(\frac{1}{3}+1\right)+...+\left(\frac{1}{2019}+1\right)\)
\(A=\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}\)
Và \(B=\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}}{\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}}=1\)
c) \(A=\frac{99}{1}+\frac{98}{2}+...+\frac{1}{99}\)
\(A=99+\frac{98}{2}+...+\frac{1}{99}\)(có 98 phân số nên ta cộng 1 vào từng phân số)
\(A=\left(\frac{98}{2}+1\right)+\left(\frac{97}{3}+1\right)+...+\left(\frac{1}{99}+1\right)+1\)
\(A=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}+1\)
\(A=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)\)
Và \(B=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\)
\(\Rightarrow\frac{A}{B}=\frac{100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}}=100\)
DL
2 tháng 6 2018
a)\(B=\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+...+\frac{100}{99}\)
\(B=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{99}\right)\)
\(\Rightarrow B=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}\right)\)
\(\Rightarrow B=98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}\)
\(\Rightarrow A:B=\frac{A}{B}=\frac{98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}}{98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}}=1.\)
Vậy \(A:B=1.\)
b)\(B=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{2019}\right)\)
\(\Rightarrow B=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right)\)
\(\Rightarrow B=2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\)
\(\Rightarrow A:B=\frac{A}{B}=\frac{2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}}{2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}}=1.\)
Vậy \(A:B=1.\)
c)\(A=\left(1+1+...+1\right)+\frac{98}{2}+\frac{97}{3}+...+\frac{2}{98}+\frac{1}{99}\)
\(A=\left(1+\frac{98}{2}\right)+\left(1+\frac{97}{3}\right)+...+\left(1+\frac{2}{98}\right)+\left(1+\frac{1}{99}\right)\)
\(A=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{98}+\frac{100}{99}\)
\(A=100\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}\right)\)
\(\Rightarrow A:B=\frac{A}{B}=\frac{100\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}}=1.\)
Vậy \(A:B=1.\)
24 tháng 2 2018
a)\(1-2+3-4+5-6+7-8+8-9+9-10\)
=\(\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+\left(7-8\right)+\left(8-9\right)+\left(9-10\right)\)
\(=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)\)
\(=\left(-1\right).6\)
\(=-6\)
b)\(1-2+3-4+...+99-100\)
\(=\left(1-2\right)+\left(3-4\right)+...+\left(99-100\right)\)}\(\left[\left(100-1\right):1+1\right]:2=50\)(cặp)
\(=\left(-1\right)+\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)} 50 số (-1)
\(=\left(-1\right).50\)
\(=-50\)
c)\(1-3+5-7+9-11+13-15\)
\(=\left(1-3\right)+\left(5-7\right)+\left(9-11\right)+\left(13-15\right)\)
\(=\left(-2\right)+\left(-2\right)+\left(-2\right)+\left(-2\right)\)
\(=\left(-2\right).4\)
\(=-8\)
d)\(1-3+5-7+...-99+101\) (Đối với bài này, có vẻ đề sai, mình đã sửa lại rồi
\(=\left(1-3\right)+\left(5-7\right)+...+\left(97-99\right)+101\) } \(\left[\left(99-1\right):2+1\right]:2=25\)(cặp)
\(=\left(-2\right)+\left(-2\right)+\left(-2\right)+...+\left(-2\right)\) } 25 số (-2)
\(=\left(-2\right).25\)
\(=-50\)
e)\(-1-2-3-4-...-99-100\)
\(=\left(-1\right)+\left(-2\right)+\left(-3\right)+...+\left(-99\right)+\left(-100\right)\)
\(=\left[\left(-1\right)+\left(-100\right)\right]+\left[\left(-2\right)+\left(-99\right)\right]+...+\left[\left(-51\right)+\left(-50\right)\right]\) } \(\left[\left(100-1\right):1+1\right]:2=50\)(cặp) (phần này của đề bài, không thay được như (-100) hoặc (-1))
\(=\left(-100\right)+\left(-100\right)+\left(-100\right)+...+\left(-100\right)\)} 50 số (-100)
\(=\left(-100\right).50\)
\(=-5000\)
NT
2
6 tháng 5 2021
A = 1*2*3 + 2*3*4 + 3*4*5 ... + 99*100*101
=> 4A = 1*2*3*4 + 2*3*4*4 + 3*4*5*4 + ... +99*100*101*4
=> 4A = 1*2*3*4 + 2*3*4*(5 - 1) + 3*4*5*( 6 - 2) + ... + 99*100*101*(102 - 98)
=> 4A = 1*2*3*4 + 2*3*4*5 - 1*2*3*4 + 3*4*5*6 - 2*3*4*5 + ... + 99*100*101*102 - 98*99*100*101
=> 4A = 99*100*101*102
=> 4A = 101989800
=> A = 25497450
P
1
S
22 tháng 4 2018
\(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}\)
\(A=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(A=\frac{1}{2}\cdot\frac{1}{4949}\)
\(A=\frac{1}{9898}\)
DD
23 tháng 6 2017
A = 1 . 2 + 2 . 3 + 3 . 4 + ... + 99 .100
3 . A = 1. 2 . 3 + 2 . 3 . 3 + 3 . 4 . 3 + ... + 99 . 100 . 3
3 . A = 1 . 2 . 3 + 2 . 3 . ( 4 - 1 ) + 3 . 4 . ( 5 - 2 ) + ... + 99 . 100 . ( 1001 - 998 )
3 . A = 1 . 2 . 3 + 2 . 3 . 4 - 1 . 2 . 3 + 3 . 4 . 5 - 2 . 3 . 4 + ... + 99 . 100 . 1001 - 998 . 99 . 100
3 . A = 99 . ( 100 . 10 )
A = ( 99 . 100 . 10 ) : 3
A = 33000
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)
đơn giản