Cho M= 53+832+1133+...+302310053+832+1133+3023100CMR...
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Cho M= CMR M
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19 tháng 4 2015
\(M=\frac{20}{112}+\frac{20}{280}+\frac{20}{520}+\frac{20}{832}\)
\(M=\frac{20}{8.14}+\frac{20}{14.20}+\frac{20}{20.26}+\frac{20}{26.32}\)
\(M=\frac{20}{6}.\left(\frac{6}{8.14}+\frac{6}{14.20}+\frac{6}{20.26}+\frac{6}{26.32}\right)\)
\(M=\frac{20}{6}\left(\frac{1}{8}-\frac{1}{14}+\frac{1}{14}-\frac{1}{20}+\frac{1}{20}-\frac{1}{26}+\frac{1}{26}-\frac{1}{32}\right)\)
\(M=\frac{20}{6}\left(\frac{1}{8}-\frac{1}{32}\right)\)
\(M=\frac{20}{6}\cdot\frac{3}{32}\)
\(M=\frac{5}{16}\)
3 tháng 5 2016
$M=\frac{20}{112}+\frac{20}{280}+\frac{20}{520}+\frac{20}{832}$
$M=\frac{20}{8.14}+\frac{20}{14.20}+\frac{20}{20.26}+\frac{20}{26.32}$
$M=\frac{20}{6}.\left(\frac{6}{8.14}+\frac{6}{14.20}+\frac{6}{20.26}+\frac{6}{26.32}\right)$
$M=\frac{20}{6}\left(\frac{1}{8}-\frac{1}{14}+\frac{1}{14}-\frac{1}{20}+\frac{1}{20}-\frac{1}{26}+\frac{1}{26}-\frac{1}{32}\right)$
$M=\frac{20}{6}\left(\frac{1}{8}-\frac{1}{32}\right)$
$M=\frac{20}{6}\cdot \frac{3}{32}$
$M=\frac{5}{16}$
3 tháng 4 2016
ta có c+3=8 vậy c=8-3=5
b+3=5 vậy b=5-3=2
a+1=2 vậy a =2-1=1
vậy abc là 125
24 tháng 4 2017
M=20/8.14+20/14.20+20/20.26+20/26.32
=>M=20/6(6/8.14+6/14.20+6/20.26+6/26.32)
=>M=20/6(1/8-1/14+1/14-1/20+1/20-1/26+1/26-1/32)
=>M=20/6(1/8-1/32)
= 20/6.3/32
=5/16
PH
21 tháng 4 2015
abc +1133=abcx10+8
1133=abcx9+8
abcx9+8=1133
abcx9=1133-8
abcx9=1125
abc=1125:9
abc=125
1 đúng nhé
21 tháng 4 2015
abc + 1133 = abc8
abc8=abc+1133
abc.10+8=abc+1133
abc.9+8=1133(đã bớt abc.1 ở cả 2 vế)
abc.9=1133-8
abc.9=1125
abc=1125:9
abc=125
=>abc=125
Dấu chấm ở đây là phép nhân nha!
LC
8 tháng 8 2015
abc+1133=abc8
=>abc+1133=abc.10+8
=>abc.10-abc=1133-8
=>abc.9=1125
=>abc=1125:9
=>abc=125
Vậy abc=125
\(M=\dfrac{5}{3}+\dfrac{8}{3^2}+...+\dfrac{302}{3^{100}}\\ 3M=5+\dfrac{8}{3}+\dfrac{11}{3^2}+...+\dfrac{302}{3^{99}}\\ 3M-M=\left(5+\dfrac{8}{3}+\dfrac{11}{3^2}+...+\dfrac{302}{3^{99}}\right)-\left(\dfrac{5}{3}+\dfrac{8}{3^2}+...+\dfrac{302}{3^{100}}\right)\\ 2M=5+\left(\dfrac{8}{3}-\dfrac{5}{3}\right)+\left(\dfrac{11}{3^2}-\dfrac{8}{3^2}\right)+....+\left(\dfrac{302}{3^{99}}-\dfrac{299}{3^{99}}\right)+\dfrac{302}{3^{100}}\\ 2M=5+1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}+\dfrac{302}{3^{100}}\\ 2M=6+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}+\dfrac{302}{3^{100}}\\ 6M=18+1+\dfrac{1}{3}+...+\dfrac{1}{3^{97}}+\dfrac{302}{3^{99}}\\ 6M-2M=\left(19+\dfrac{1}{3}+...+\dfrac{1}{3^{97}}+\dfrac{302}{3^{99}}\right)-\left(6+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}+\dfrac{302}{3^{100}}\right)\\ 4M=13-\dfrac{1}{3^{98}}+\dfrac{302}{3^{99}}-\dfrac{302}{3^{100}}\\ 4M=13+\dfrac{595}{3^{100}}\\ M=\dfrac{1}{4}\left(13+\dfrac{595}{3^{100}}\right)=\dfrac{13}{4}+\dfrac{1}{4}\cdot\dfrac{595}{3^{100}}\)
Vì:
\(595< 3^{100}=>\dfrac{595}{3^{100}}< 1=>\dfrac{1}{4}\cdot\dfrac{595}{3^{100}}< \dfrac{1}{4}\\ =>M=\dfrac{13}{4}+\dfrac{1}{4}\cdot\dfrac{595}{3^{100}}< \dfrac{13}{4}+\dfrac{1}{4}=\dfrac{7}{2}=3\dfrac{1}{2}\)