\(p=\frac{1.2.4+2.3.5+3.4.6+...+100.101.103}{1.2^2+2.3^2+3.4^2+...+100.101^2}\)Rút gọn P
K
Khách
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TP
0
PV
4
3 tháng 5 2018
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{100^2}{100.101}\)
\(=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}...\frac{100.100}{100.101}\)
\(=\frac{1.1.2.2.3.3...100.100}{1.2.2.3.3.4...100.101}\)
\(=\frac{\left(1.2.3...100\right).\left(1.2.3...100\right)}{\left(1.2.3....100\right).\left(2.3.4...101\right)}\)
\(=\frac{1.1}{1.101}\)
\(=\frac{1}{101}\)
WH
3 tháng 5 2018
\(\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}.....\frac{100^2}{100\cdot101}\)
\(=\frac{1.1}{1\cdot2}\cdot\frac{2.2}{2.3}\cdot\frac{3.3}{3.4}.....\frac{100.100}{100.101}\)
\(=\frac{\left(1\cdot2\cdot3\cdot\cdot\cdot\cdot\cdot\cdot100\right)\left(1\cdot2\cdot3\cdot\cdot\cdot\cdot\cdot100\right)}{\left(1\cdot2\cdot3\cdot4\cdot\cdot\cdot\cdot\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot\cdot\cdot\cdot\cdot101\right)}\)
\(=\frac{1}{101}\)
S
1
2 tháng 8 2019
\(\frac{1.1}{1.2}.\frac{2.2}{2.3}\frac{3.3}{3.4}...\frac{100.100}{100.101}\)
\(=\frac{\left(1.2.3...100\right).\left(1.2.3...100\right)}{\left(1.2.3...100\right).\left(2.3...101\right)}\)
\(=\frac{1}{1.101}\)
\(=\frac{1}{101}\)
k cho mk nha
H
2
PT
11 tháng 5 2015
=2(1/1.2+1/2.3+...+1/100.101)
=2(1/1-1/2+1/2-...+1/100-1/101)
=2(1-1/101)
=2.100/101
=200/101
PG
0
LK
2
11 tháng 4 2017
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.......\frac{99^2}{99.100}.\frac{100^2}{100.101}\)
\(=\frac{1.2.3.....100}{1.2.3....100}.\frac{1.2.3....100}{2.3.4...101}\)
\(=1.\frac{1}{101}=\frac{1}{101}\)
11 tháng 4 2017
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{99^2}{99.100}.\frac{100^2}{100.101}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}.\frac{100}{101}\)
\(=\frac{1.2.3...99.100}{2.3.4...100.101}\)
\(=\frac{1}{101}\)
NN
1.Tính
A= (1-1/22).(1-1/32)...(1-1/1002)
B= -1/1.2-1/2.3-1/3.4-...-1/100.101
C= 1.2+2.3+3.4+...+100.101
2
13 tháng 9 2016
Lời giải :
Đặt S=1.2+2.3+3.4+4.5+…+99.100+100.101
3S=1.2.3+2.3.3+3.4.3+4.5.3+…+99.100.3+100.101.3
=1.2(3−0)+2.3(4−1)+3.4(5−2)+4.5(6−3)+…+99.100(101−98)+100.101(102−99)
=0.1.2-1.2.3+1.2.3-2.3.4+...+99.100.101-100.101.102
=100.101.102
S=100.101.34=343400
12 tháng 10 2022
1.Tính
a) Ta có:
A=(1-1/22).(1-1/32)...(1-1/1002)
=>A=3/22.8/32.....9999/1002
=>A=(1.3/2.2).(2.4/3.3).....(99.101/100.100)
=>A=(1.2.3.....99/2.3.4.....100).(3.4.5.....101/2.3.4.....100)
=>A=1/100.101/2
=>A=101/200
b) Ta có:
B=-1/1.2-1/2.3-1/3.4-...-1/100.101
=>B=-(1/1.2+1/2.3+1/3.4+...+1/100.101)
=>B=-(1-1/2+1/2-1/3+1/3-1/4+...+1/100-1/101)
=>B=-(1-1/101)
=>B=-100/101
c) Ta có:
C=1.2+2.3+3.4+...+100.101
=>3C=1.2.3+2.3.3+3.4.3+...+100.101.3
=>3C=1.2.3+2.3.(4-1)+3.4.(5-2)+...+100.101.(102-99)
=>3C=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+...+100.101.102
=>3C=100.101.102
=>3C=1030200
=>C=343400
Chúc bạn hok tốt nhé >:)!!!!!
PS
0
Tử số = \(1.2.4+2.3.5+3.4.6+...+100.101.103\)
\(=1.2.\left(3+1\right)+2.3.\left(4+1\right)+3.4.\left(5+1\right)+...+100.101.\left(102+1\right)\)
\(=1.2.3+1.2+2.3.4+2.3+3.4.5+3.4+...+100.101.102+100.101\)
\(=\left(1.2.3+2.3.4+3.4.5+...+100.101.102\right)+\left(1.2+2.3+3.4+...+100.101\right)\)
Mẫu số = \(1.2^2+2.3^2+3.4^2+...+100.101^2\)
\(=1.2.\left(3-1\right)+2.3.\left(4-1\right)+3.4.\left(5-1\right)+...+100.101.\left(102-1\right)\)
\(=1.2.3-1.2+2.3.4-2.3+3.4.5-3.4+...+100.101.102-100.101\)
\(=\left(1.2.3+2.3.4+3.4.5+...+100.101.102\right)-\left(1.2+2.3+3.4+...+100.101\right)\)
đặt \(A=1.2.3+2.3.4+3.4.5+...+100.101.102\) và \(B=1.2+2.3+3.4+...+100.101\)
bạn tự tính : \(A=\frac{100.101.102.103}{4}=25.101.102.103\); \(B=\frac{100.101.102}{3}=100.101.34\)
rồi thay vào tìm P=\(\frac{A+B}{A-B}\)