A=\(\frac{15}{90.94}+\frac{15}{94.98}+.....+\frac{15}{146.150}\)
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LV
3
ML
30 tháng 6 2017
\(\frac{15}{90.94}+\frac{15}{94.98}+\frac{15}{98.102}+...+\frac{15}{146.150}\)
\(=\frac{15}{4}\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+\frac{1}{98}-\frac{1}{102}+...+\frac{1}{146}-\frac{1}{150}\right)\)
\(=\frac{15}{4}\left(\frac{1}{90}-\frac{1}{150}\right)\)
\(=\frac{15}{4}.\frac{1}{225}\)
\(=\frac{1}{60}\)
DP
30 tháng 6 2017
\(\frac{15}{90\cdot94}+\frac{15}{94\cdot98}+\frac{15}{98\cdot102}+...+\frac{15}{146\cdot150}\)
\(=\frac{15}{4}\cdot\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+\frac{1}{98}-\frac{1}{102}+...+\frac{1}{146}-\frac{1}{150}\right)\)
\(=\frac{15}{4}\cdot\left(\frac{1}{90}-\frac{1}{150}\right)\)
\(=\frac{15}{4}\cdot\frac{1}{225}=\frac{1}{60}\)
DA
6
8 tháng 5 2017
\(\dfrac{15}{90\cdot94}+\dfrac{15}{94\cdot98}+\dfrac{15}{98\cdot102}+...+\dfrac{15}{146\cdot150}\)
\(=\dfrac{15}{4}\left(\dfrac{4}{90\cdot94}+\dfrac{4}{94\cdot98}+...+\dfrac{4}{146\cdot150}\right)\)
\(=\dfrac{15}{4}\left(\dfrac{1}{90}-\dfrac{1}{94}+\dfrac{1}{94}-\dfrac{1}{98}+...+\dfrac{1}{146}-\dfrac{1}{150}\right)\)
\(=\dfrac{15}{4}\left(\dfrac{1}{90}-\dfrac{1}{150}\right)=\dfrac{15}{4}\cdot\dfrac{1}{225}=\dfrac{1}{60}\)
DV
5 tháng 3 2017
15.1/5(1/90.94+1/84.98+...+1/146.150)
=3(1/90-1/94+1/94-1/98+...+1/146-1/150)
=3(1/90-1/150)
3.1/225
=1/75
MA
11 tháng 9 2016
\(A=\frac{15}{90.94}+\frac{15}{94.98}+...+\frac{15}{146.150}\)
\(A=\frac{4}{4}\left(\frac{15}{90.94}+\frac{15}{94.98}+...+\frac{15}{146.150}\right)\)
\(A=\frac{15}{4}\left(\frac{4}{90.94}+\frac{4}{94.98}+...+\frac{4}{146.150}\right)\)
\(A=\frac{15}{4}\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+...+\frac{1}{146}-\frac{1}{150}\right)\)
\(A=\frac{15}{4}\left(\frac{1}{90}-\frac{1}{150}\right)\)
\(A=\frac{15}{4}.\frac{1}{225}=\frac{1}{60}\)
\(B=\frac{6}{15.18}+\frac{6}{18.21}+...+\frac{6}{87.90}\)
\(B=\frac{3}{3}\left(\frac{6}{15.18}+\frac{6}{18.21}+...+\frac{6}{87.90}\right)\)
\(B=2\left(\frac{3}{15.18}+\frac{3}{18.21}+...+\frac{3}{87.90}\right)\)
\(B=2\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)
\(B=2\left(\frac{1}{15}-\frac{1}{90}\right)\)
\(B=2.\frac{1}{18}=\frac{1}{9}\)
15 tháng 7 2020
Trả lời:
\(A=\frac{15}{90.94}+\frac{15}{94.98}+...+\frac{15}{146.150}\)
\(A=\frac{15}{4}.\left(\frac{4}{90.94}+\frac{4}{94.98}+...+\frac{4}{146.150}\right)\)
\(A=\frac{15}{4}.\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+...+\frac{1}{146}-\frac{1}{150}\right)\)
\(A=\frac{15}{4}.\left(\frac{1}{90}-\frac{1}{150}\right)\)
\(A=\frac{15}{4}.\frac{1}{225}\)
\(A=\frac{1}{60}\)
\(B=\frac{6}{15.18}+\frac{6}{18.21}+...+\frac{6}{87.90}\)
\(B=2.\left(\frac{3}{15.18}+\frac{3}{18.21}+...+\frac{3}{87.90}\right)\)
\(B=2.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)
\(B=2.\left(\frac{1}{15}-\frac{1}{90}\right)\)
\(B=2.\frac{1}{18}\)
\(B=\frac{1}{9}\)
20 tháng 3 2018
Ta có :
\(H=\frac{15}{90.94}+\frac{15}{94.98}+\frac{15}{98.102}+...+\frac{15}{146.150}\)
\(H=\frac{15}{4}\left(\frac{4}{90.94}+\frac{4}{94.98}+\frac{4}{98.102}+...+\frac{4}{146.150}\right)\)
\(H=\frac{15}{4}\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+\frac{1}{98}-\frac{1}{102}+...+\frac{1}{146}-\frac{1}{150}\right)\)
\(H=\frac{15}{4}\left(\frac{1}{90}-\frac{1}{150}\right)\)
\(H=\frac{15}{4}.\frac{1}{225}\)
\(H=\frac{1}{60}\)
Vậy \(H=\frac{1}{60}\)
Chúc bạn học tốt ~
S
20 tháng 3 2018
\(H=\frac{15}{90\cdot94}+\frac{15}{94\cdot98}+\frac{15}{98\cdot102}+...+\frac{15}{146\cdot150}\)
\(H=15\left(\frac{1}{90\cdot94}+\frac{1}{94\cdot98}+\frac{1}{98\cdot102}+...+\frac{1}{146\cdot150}\right)\)
\(H=15\left[\frac{1}{4}\left(\frac{4}{90\cdot94}+\frac{4}{94\cdot98}+\frac{4}{98\cdot102}+...+\frac{4}{146\cdot150}\right)\right]\)
\(H=15\left[\frac{1}{4}\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+\frac{1}{98}-\frac{1}{102}+...+\frac{1}{146}-\frac{1}{150}\right)\right]\)
\(H=15\left[\frac{1}{4}\left(\frac{1}{90}-\frac{1}{150}\right)\right]\)
\(H=15\left[\frac{1}{4}\cdot\frac{1}{225}\right]\)
\(H=15\cdot\frac{1}{900}\)
\(H=\frac{1}{60}\)
DP
5 tháng 7 2017
\(E=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+...+\frac{1}{73\cdot75}\)
\(E=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)
\(\Rightarrow E=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)=\frac{1}{2}\cdot\frac{2}{75}=\frac{1}{75}\)
\(F=\frac{15}{90\cdot94}+\frac{15}{94\cdot98}+...+\frac{15}{146\cdot150}\)
\(F=\frac{15}{4}\cdot\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+...+\frac{1}{146}-\frac{1}{150}\right)\)
\(\Rightarrow F=\frac{15}{4}\cdot\left(\frac{1}{90}-\frac{1}{150}\right)=\frac{15}{4}\cdot\frac{1}{225}=\frac{1}{60}\)
\(G=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
\(G=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(G=\frac{5}{4\cdot7}+\frac{5}{7\cdot10}+\frac{5}{10\cdot13}+...+\frac{5}{25\cdot28}\)
\(G=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(\Rightarrow G=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{5}{3}\cdot\frac{3}{14}=\frac{5}{14}\)
25 tháng 6 2016
a ) 15 / 90 x 94 + 15 / 94 x 98 + 15 / 98 x102+...+ 15 / 146 x 150
= 15/4 x ( 4/90 x 94 + 4/94 x 98 + ... + 4/ 146 x 150 )
= 15/4 x ( 1/90 - 1/94 + 1/94 - 1/98 + ... + 1/146 - 1/150 )
= 15/4 x ( 1/90 - 1/150 )
= 15/4 x 2/450
= 1/60
b ) 10 / 56 + 10/ 140 + 10 / 260+...+ 10 /1400
= 5/28 + 5/70 + 5/130 + ... + 5/700
= 5/4 x 7 + 5/7 x 10 + 5/10 x 13 + ... + 5 /25 x 28
= 5/3 x ( 3/4 x 7 + 3/7 x 10 + ... + 3/25 x 28 )
= 5/3 x ( 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/25 - 1/28 )
= 5/3 x ( 1/4 - 1/28 )
= 5/3 x 6/28
= 5/14
1 tháng 7 2016
ngài kiệt lí đây gian lận tuyệt vời.
vĩ đại như nước trong nguồn chảy ra.
\(A=\frac{15}{90.94}+\frac{15}{94.98}+...+\frac{15}{146.150}\)
\(A=15\left(\frac{1}{90.94}+\frac{1}{94.98}+...+\frac{1}{146.150}\right)\)
\(A=\frac{15}{4}\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+...+\frac{1}{146}-\frac{1}{150}\right)\)
\(A=\frac{15}{4}\left(\frac{1}{90}-\frac{1}{150}\right)\)
\(A=\frac{15}{4}.\frac{1}{225}\)
\(A=\frac{1}{60}\)