Tìm y
y x ( 2/1x2 + 2 / 2x3 + 2 / 3x4 +... + 2/99x100 ) = 198/100
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HV
0
BT
11 tháng 3 2017
A=2(\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\))=2(\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\))
=> A=2(\(\frac{1}{1}-\frac{1}{100}\))=2.\(\frac{99}{100}=\frac{99}{50}\)
ĐS: A=99/50
11 tháng 3 2017
\(\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+\frac{2}{4\times5}+...+\frac{2}{99\times100}\)
\(=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{99\times100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{1}-\frac{1}{100}\)
\(=\frac{99}{100}\)
LL
0
NA
1
HT
27 tháng 10 2015
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{99.100}\)
= \(2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\right)\)
= \(2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\right)\)
= \(2.\left(1-\frac{1}{100}\right)\)
= \(2.\frac{99}{100}\)
= \(\frac{99}{50}\)
NT
2
10 tháng 9 2017
Đặt \(A=\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{99.100}\)
\(A=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(A=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=2.\left(1-\frac{1}{100}\right)\)
\(A=\frac{2.99}{100}\)
\(A=\frac{99}{50}=1\frac{49}{50}\)
10 tháng 9 2017
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2\left(1-\frac{1}{100}\right)=2.\frac{99}{100}\)
\(=\frac{99}{50}\)
DD
Đoàn Đức Hà
Giáo viên
25 tháng 5 2021
b) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}\)
\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)
\(=\frac{1}{2}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{2015-2013}{2013.2015}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{2015}\right)=\frac{1007}{2015}\)
Phương trình tương đương với:
\(\frac{1007X}{2015}=\frac{4}{2015}\Leftrightarrow X=\frac{4}{1007}\)
DD
Đoàn Đức Hà
Giáo viên
25 tháng 5 2021
c) \(\frac{x+1}{2015}+\frac{x+2}{2016}=\frac{x+3}{2017}+\frac{x+4}{2018}\)
\(\Leftrightarrow\frac{x+1}{2015}-1+\frac{x+2}{2016}-1=\frac{x+3}{2017}-1+\frac{x+4}{2018}-1\)
\(\Leftrightarrow\frac{x-2014}{2015}+\frac{x-2014}{2016}=\frac{x-2014}{2017}+\frac{x-2014}{2018}\)
\(\Leftrightarrow x-2014=0\)
\(\Leftrightarrow x=2014\)
15 tháng 7 2017
Ta có : P = 1.2.2 + 2.3.3 + ....+ 99.100.100
=1.2.(3 - 1) + 2.3.(4 - 1) + ....+99.100.(101 - 1)
= (1.2.3 + 2.3.4 + .... + 99.100.101) - (2.3 + 3.4+.....+99.100)
Đặt B = 1.2.3 + 2.3.4 + 4.5.6 +...+ 99.100.101
4B = 1.2.3.(4 - 0)+2.3.4.(5 - 1) + ... + (99.100.101(102 - 98)
4B = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 +...+ 99.100.101.102 - 98.99.100.101
4B = 99.100.101.102
4B = 101989800
B = 25497450
Đặt C = 1.2 + 2.3 + 3.4 +...+ 99.100
3C = 1.2.(3 - 0) + 2.3.(4 - 1) +...+ 99.100.(101 - 98)
3C = 1.2.3 + 2.3.4 - 1.2.3 +...+ 99.100.101 - 98.99.100
3C = 99.100.101
3C = 999900
C = 999900 : 3
C = 333300
Vậy: P = 25497450 – 333300 = 25164150
CM
12 tháng 9 2017
A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
A x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
A x 3 = 99x100x101
A = 99x100x101 : 3
A = 333300
16 tháng 11 2021
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gấp lắm ạ giúp em với
\(\Leftrightarrow y\cdot\dfrac{99}{50}=\dfrac{198}{100}=\dfrac{99}{50}\)
hay y=1