LT
2
K
Khách
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LN
1
11 tháng 10 2019
ta có : 1.2+2.3+3.4+.....+99.100=99.100.101 /3 =333300
mà 1.2+2.3+....+9.10+9.10.11/3=330
=>E= 333300-330=332970
TQ
1
LN
3
29 tháng 6 2015
\(A=\frac{1}{7}.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{68}-\frac{1}{70}\right)\)
\(A=\frac{1}{7}.\left(\frac{1}{10}-\frac{1}{70}\right)=\frac{1}{7}.\frac{3}{35}=\frac{3}{245}\)
29 tháng 6 2015
A=\(\frac{7}{10.11}\)+\(\frac{7}{11.12}\)+\(\frac{7}{12.13}\)+...+\(\frac{7}{69.70}\)
A=\(\frac{7}{10}\)-\(\frac{7}{11}\)+\(\frac{7}{11}\)-\(\frac{7}{12}\)+\(\frac{7}{12}\)-\(\frac{7}{13}\)+...+\(\frac{7}{69}\)-\(\frac{7}{70}\)
A=\(\frac{7}{10}-\frac{7}{70}\)
A=\(\frac{7}{10}-\frac{1}{10}\)
Ạ=\(\frac{6}{10}=\frac{3}{5}\).
LT
1
18 tháng 8 2016
\(2\left(\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1}{9}\)
\(\left(\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1}{9}.\frac{1}{2}\)
\(\frac{1}{9}-\frac{1}{x+1}=\frac{1}{18}\)
\(\frac{1}{x+1}=\frac{1}{9}-\frac{1}{18}=\frac{1}{9}\)
=>x+1=9
=>x=8
thanks bạn
21 tháng 7 2018
C=1/2.(3/60.63+....+3/117.120)+1/1003
C=1/2.(1/60-1/63+....+1/117-1/120)+1/1003
....còn lại tự làm nha, bài còn lại cũng tương tự
Y
21 tháng 7 2018
Ta có:
\(C=\dfrac{2}{60.63}+\dfrac{2}{63.66}+...+\dfrac{2}{117.120}+\dfrac{2}{2006}\)
\(C=2\left(\dfrac{1}{60.63}+\dfrac{1}{63.66}+...+\dfrac{1}{117.120}\right)+\dfrac{2}{2006}\)
\(C=2.\dfrac{1}{3}\left(\dfrac{3}{60.63}+\dfrac{3}{63.66}+...+\dfrac{3}{117.120}\right)+\dfrac{2}{2006}\)
\(C=\dfrac{2}{3}\left(\dfrac{1}{60}-\dfrac{1}{63}+\dfrac{1}{63}-\dfrac{1}{66}+...+\dfrac{1}{117}-\dfrac{1}{120}\right)+\dfrac{2}{2006}\)
\(C=\dfrac{2}{3}\left(\dfrac{1}{60}-\dfrac{1}{120}\right)+\dfrac{2}{2006}\)
\(C=\dfrac{2}{3}.\dfrac{1}{120}+\dfrac{2}{2006}\)
\(C=\dfrac{1}{180}+\dfrac{2}{2006}\)
Ta lại có:
\(D=\dfrac{5}{40.44}+\dfrac{5}{44.48}+...+\dfrac{5}{76.80}+\dfrac{5}{2006}\)
\(D=5\left(\dfrac{1}{40.44}+\dfrac{1}{44.48}+...+\dfrac{1}{76.80}\right)+\dfrac{5}{2006}\)
\(D=5.\dfrac{1}{4}\left(\dfrac{4}{40.44}+\dfrac{4}{44.48}+...+\dfrac{4}{76.80}\right)+\dfrac{5}{2006}\)
\(D=\dfrac{5}{4}\left(\dfrac{1}{40}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{48}+...+\dfrac{1}{76}-\dfrac{1}{80}\right)+\dfrac{5}{2006}\)
\(D=\dfrac{5}{4}\left(\dfrac{1}{40}-\dfrac{1}{80}\right)+\dfrac{5}{2006}\)
\(D=\dfrac{5}{4}.\dfrac{1}{80}+\dfrac{5}{2006}\)
\(D=\dfrac{1}{64}+\dfrac{5}{2006}\)
Vì \(\dfrac{1}{180}< \dfrac{1}{64}\)
\(\dfrac{2}{2006}< \dfrac{5}{2006}\)
\(\Rightarrow\dfrac{1}{180}+\dfrac{2}{2006}< \dfrac{1}{64}+\dfrac{5}{2006}\)
\(\Rightarrow C< D\)
BH
21 tháng 7 2018
dở ẹt nhu cu net ma ko biet lamb tao hoc lop mau giao tao cung biet tra loi dung la ngu
29 tháng 8 2016
mình biết
\(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{2015.2016}\)
\(\Rightarrow\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{2015}-\frac{1}{2016}\)
ta rút gọn được
\(\Rightarrow\frac{1}{10}-\frac{1}{2016}\)
\(\Rightarrow\frac{1003}{10080}\)
F
20 tháng 4 2019
\(A=\frac{2}{60\cdot63}+\frac{2}{63\cdot66}+...+\frac{2}{117\cdot120}+\frac{2}{2003}\)
\(\text{Đặt }C=\frac{2}{60\cdot63}+\frac{2}{63\cdot66}+...+\frac{2}{117\cdot120}\)
\(C=\frac{2}{3}\left(\frac{3}{60\cdot63}+\frac{3}{63\cdot66}+...+\frac{3}{117\cdot120}\right)\)
\(C=\frac{2}{3}\left(\frac{1}{60}-\frac{1}{63}+\frac{1}{63}-\frac{1}{66}+...+\frac{1}{117}-\frac{1}{120}\right)\)
\(C=\frac{2}{3}\left(\frac{1}{60}-\frac{1}{120}\right)\)
\(C=\frac{2}{3}\cdot\frac{1}{120}\)
\(C=\frac{1}{180}\)
\(\text{Thay }C=\frac{1}{180}\text{Ta có : }\) \(A=\frac{1}{180}+\frac{2}{2003}\)
\(B=\frac{5}{40\cdot44}+\frac{5}{44\cdot48}+...+\frac{5}{76\cdot80}+\frac{5}{2003}\)
\(\text{Đặt }D=\frac{5}{40\cdot44}+\frac{5}{44\cdot48}+...+\frac{5}{76\cdot80}\)
\(D=\frac{5}{4}\left(\frac{4}{40\cdot44}+\frac{4}{44\cdot48}+...+\frac{4}{76\cdot80}\right)\)
\(D=\frac{5}{4}\left(\frac{1}{40}-\frac{1}{44}+\frac{1}{44}-\frac{1}{48}+...+\frac{1}{76}-\frac{1}{80}\right)\)
\(D=\frac{5}{4}\left(\frac{1}{40}-\frac{1}{80}\right)\)
\(D=\frac{5}{4}\cdot\frac{1}{80}\)
\(D=\frac{1}{64}\)
\(\text{Thay }D=\frac{1}{64}\text{ Ta có : }B=\frac{1}{64}+\frac{5}{2003}\)
\(\text{Vì }A=\frac{1}{180}+\frac{2}{2003}\text{ , }B=\frac{1}{64}+\frac{5}{2003}\)
\(\text{Có : }\frac{1}{180}< \frac{1}{64}\)
\(\frac{2}{2003}< \frac{5}{2003}\)
\(\Rightarrow\text{ }A< B\)
15*27-15*27
11.12 + 22.24 + 44.48
33.36 + 66.72 + 132.144
= 11.12 + 2.11.12.2 + 4.11.12.4
3.11.12.3 + 6.11.12.6 + 12.11.12.12
= 11.12. (1 + 2^2 + 4^2)
11.12. (3^2 + 6^2 + 12^2)
= 1 + 2^2 + 4^2
3^2. (1 + 2^2 + 4^2)
= 1
9