C=1/1.4+1/2.6+1/3.8+...+1/8.18+1/9.20
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2 tháng 7 2015
B=1/1x4+1/4x3+1/3x8+...+1/7x16+1/16x9+1/9x20
2B=2x(1/4+1/12+1/24+...+1/112+1/144+1/180
2B=2/8+2/24+2/48+...+2/224+2/288+2/360
2B=2/2x4+2/4x6+2/6x8+...+2/14x16+2/16x18+2/18x20
2B=1/2-1/4+1/4-1/6+1/6-1/8+...+1/14-1/16+1/16-1/18+1/18-1/20
2B=1/2-1/20
2B=9/20
B=9/20:2
B=9/40
TP
2
TM
0
CT
0
TL
1
MV
1 tháng 5 2017
\(A=\dfrac{3}{1\cdot4}+\dfrac{3}{2\cdot6}+\dfrac{3}{3\cdot8}+...+\dfrac{1}{2012\cdot1342}\\ =\dfrac{3}{1\cdot4}+\dfrac{3}{2\cdot6}+\dfrac{3}{3\cdot8}+...+\dfrac{3}{2012\cdot4026}\\ =\dfrac{6}{2\cdot4}+\dfrac{6}{4\cdot6}+\dfrac{6}{6\cdot8}+...+\dfrac{6}{4024\cdot4026}\\ =3\cdot\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{4024\cdot4026}\right)\\ =3\cdot\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{4024}-\dfrac{1}{4026}\right)\\ =3\cdot\left(\dfrac{1}{2}-\dfrac{1}{4026}\right)\\ =3\cdot\dfrac{1}{2}-3\cdot\dfrac{1}{4026}\\ =1,5-\dfrac{3}{4026}< 1,5\)
Vậy \(A< 1,5\left(đpcm\right)\)
8 tháng 5 2018
\(\frac{3}{1.4}+\frac{3}{2.6}+\frac{3}{3.8}+...+\frac{1}{2012.1342}\)
\(=\frac{3}{1.4}+\frac{3}{2.6}+\frac{3}{3.8}+...+\frac{3}{2012.4026}\)
\(=\frac{6}{2.4}+\frac{6}{4.6}+\frac{6}{4.8}+...+\frac{6}{4024.4026}\)
\(=3\cdot\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{4024.4026}\right)\)
\(=3\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{4024}-\frac{1}{4026}\right)\)
\(=3\cdot\left(\frac{1}{2}-\frac{1}{4026}\right)\)
\(=3\cdot\frac{1}{2}-3\cdot\frac{1}{4026}\)
\(=1,5-\frac{3}{4026}< 1,5\)
11 tháng 5 2017
\(A=\dfrac{3}{1.4}+\dfrac{3}{2.6}+\dfrac{3}{3.8}+...............+\dfrac{1}{2012.1342}\)
\(A=\dfrac{3}{1.4}+\dfrac{3}{2.6}+\dfrac{3}{3.8}+...........................+\dfrac{3}{2012.4026}\)
\(A=\dfrac{6}{2.4}+\dfrac{6}{4.6}+\dfrac{6}{6.8}+..........................+\dfrac{6}{4024.4026}\)
\(A=3\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...................+\dfrac{2}{4024.4026}\right)\)
\(A=3\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+....................+\dfrac{1}{4024}-\dfrac{1}{4026}\right)\)
\(A=3\left(\dfrac{1}{2}-\dfrac{1}{4026}\right)\)
\(A=3.\dfrac{1}{2}-3.\dfrac{1}{4026}\)
\(A=1,5-\dfrac{3}{4026}< 1,5\)
DT
Đỗ Thanh Hải
CTVVIP
11 tháng 5 2017
Ta có
A = \(\dfrac{3}{1.4}\) + \(\dfrac{3}{2.6}\) + \(\dfrac{3}{3.8}\) + ... + \(\dfrac{1}{2012.1342}\)
A = \(\dfrac{3}{1.4}\) + \(\dfrac{3}{2.6}\) + \(\dfrac{3}{3.8}\) + ... + \(\dfrac{3}{2012.4026}\)
A = \(\dfrac{6}{2.4}\) + \(\dfrac{6}{4.6}\) + \(\dfrac{6}{6.8}\) + ... + \(\dfrac{6}{4024.4026}\)
A = \(3\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{4024.4026}\right)\)
A = \(3\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{4024}-\dfrac{1}{4026}\right)\)
A = \(3\left(\dfrac{1}{2}-\dfrac{1}{4026}\right)\)
A = 3.\(\dfrac{1}{2}\) - 3.\(\dfrac{1}{4026}\)
A = 1,5 - \(3.\dfrac{1}{4026}\) < 1,5
=> A < 1,5
=> đpcm
\(C=\dfrac{1}{1\text{x}4}+\dfrac{1}{2\text{x}6}+...+\dfrac{1}{8\text{x}18}+\dfrac{1}{9\text{x}20}\)
\(=\dfrac{1}{4}\text{x}\left(1+\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{36}+\dfrac{1}{45}\right)\)
\(=\dfrac{1}{4}\text{x}\left(\dfrac{2}{2}+\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{72}+\dfrac{2}{90}\right)\)
\(=\dfrac{1}{2}\text{x}\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{72}+\dfrac{1}{90}\right)\)
\(=\dfrac{1}{2}\text{x}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(=\dfrac{1}{2}\text{x}\left(1-\dfrac{1}{10}\right)=\dfrac{1}{2}\text{x}\dfrac{9}{10}=\dfrac{9}{20}\)
Cú tui