a) Tính A=1/10+1/40+1/88+1/154+1/238+1/340
b) so sánh: 200410 + 20049 và 200510
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LT
1
28 tháng 7 2015
\(S=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{184}+\frac{1}{238}+\frac{1}{340}=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{1}{3}.\frac{9}{20}=\frac{3}{20}>\frac{2}{20}=\frac{1}{10}=0,1\)
vậy S>0,1
NP
2
HT
28 tháng 7 2015
S = \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)
S = \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)
S = \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{17}-\frac{1}{20}\right)\)
S = \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{1}{3}.\frac{9}{20}\)
S = \(\frac{3}{20}\)
S = 0,15 > 0,1
24 tháng 4 2021
Giải:
A=1/10+1/40+1/88+1/154+1/238+1/340
A=1/2.5+1/5.8+1/8.11+1/11.14+1/14.17+1/17.20
A=1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14+1/14-1/17+1/17-1/20
A=1/2-1/20
A=9/20
D=1/3+1/6+1/12+1/24+1/48
D=1/3+1/2.3+1/3.4+1/4.6+1/6.8
D=1/3+1/2-1/3+1/3-1/4+1/2.(2/4.6+2/6.8)
D=1/3+1/2-1/4+1/2.(1/4-1/6+1/6-1/8)
D=1/3+1/4+1/2.(1/4-1/8)
D=1/3+1/4+1/2.1/8
D=1/3+1/4+1/16
D=31/48
F=0,5-1/3-0,4-4/7-1/6+4/35-1/41
F=1/2-1/3-2/5-4/7-1/6+4/35-1/41
F=1/6-(-6/35)-1/6+4/35-1/41
F=(1/6-1/6)+(6/35+4/35)-1/41
F=0+2/7-1/41
F=2/7+1/41
F=75/287
Chúc bạn học tốt!
AX
1
6 tháng 3 2016
\(A=\frac{9}{10}+\frac{39}{40}+\frac{87}{88}+\frac{153}{154}+\frac{237}{238}+\frac{339}{340}=\frac{117}{20}\)
\(suyra:A<1\)
TN
30 tháng 4 2016
\(A=\frac{3}{3}.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\right)\)
\(A=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{3}{20}\)
21 tháng 1 2021
Đặt \(A=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)
=> \(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\) (dấu . có nghĩa là nhân)
=> \(3A=3\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\right)\)
\(=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}\)
\(=\frac{9}{20}\)
Đây là kiến thức lớp 6 nhá =)) bạn mà có chỗ nào ko hiểu thì hỏi ng thầy cô giạy bạn ý
VD
18 tháng 2 2018
a, \(A=\frac{1}{10}+\frac{1}{40}+...+\frac{1}{340}\)
\(\Leftrightarrow A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{17.20}\)
\(\Leftrightarrow A=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+....+\frac{3}{17.20}\right)\)
\(\Leftrightarrow A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(\Leftrightarrow A=\frac{1}{6}-\frac{1}{60}=\frac{3}{20}\)
b, \(2004^{10}+2004^9=2004^9\left(2014+1\right)=2014^9+2005\)
\(2015^{10}=2015^9.2015\)
-Vậy: \(2004^{10}+2004^9< 2005^{10}\)
LK
1
a) A = 1/2.5 + 1/5.8 + 1/8.11 + 1/11.14 + 1/14.17 + 1/17.20
=> 3A = 1/2 - 1/5 + 1/5 - .... + 1/14 - 1/17 + 1/17 - 1/20
=> 3A = 1/2 - 1/20 = 9/20
=> A = 3/20
b) 200410 + 20049 = 20049(1+2004) = 20049 . 2005
200510 = 20059 . 2005
Do 20059 > 20049 nên 200410 + 20049 < 200510