so sánh 99!/1 với 50^99/1
Cho B = 1/50 + 1/51 + 1/52 + .....+ 1/99.So sánh B với 1
B=1/50+1/51+1/52+...+1/99
Ta có: 1/50=1/50
1/51<1/50
1/52<1/50
..............
1/99<1/50
1/50+1/51+1/52+...+1/99<1/50+1/50+1/50+...+1/50(50 phân số 1/50)
B<1
Cho M = 1/50 + 1/51 + ... + 1/98 + 1/99 . So sánh M với 1/2
Ta có :
\(\frac{1}{50}>\frac{1}{100}\)
\(\frac{1}{51}>\frac{1}{100}\)
............
\(\frac{1}{98}>\frac{1}{100}\)
\(\frac{1}{99}>\frac{1}{100}\)
\(\Rightarrow\frac{1}{50}+\frac{1}{51}+....+\frac{1}{98}+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+....+\frac{1}{100}=\frac{50.1}{100}=\frac{1}{2}\)
\(\Rightarrow M>\frac{1}{2}\)
Ta có: \(\frac{1}{50}>\frac{1}{51}>....>\frac{1}{99}\)
\(\Rightarrow M>\frac{1}{99}+\frac{1}{99}+...+\frac{1}{99}=\frac{50}{99}>\frac{50}{100}=\frac{1}{2}\)
Vậy M > 1/2
Ta có :
M = \(\frac{1}{50}+\frac{1}{51}+...+\frac{1}{98}+\frac{1}{99}\)
M > \(\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}+\frac{1}{100}\)( 50 số hạng )
\(=\frac{1}{100}.50=\frac{1}{2}\)
Vậy M > \(\frac{1}{2}\)
cho s = 1/50 + 1/51 + 1/52 + 1/53 + .......... + 1/99 + 1/100 . hãy so sánh s với 5/6 cứu mình với
so sánh 1+1/2+1/3+...+1/2^99 và 50
Cho B=\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^{99}}\) ,So sánh B với 50
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? crisdevergamemer
1/1*1+1/2*3+1/3*5+1/4*7+...+1/49*97+1/50*99. Hãy so sánh với 7/6
cho A=1/11+1/12+1/13+1/14+...+1/50
so sánh A với 1/2
cho B=1/50+1/51+1/52+...+1/98+1/99
chứng minh rằng b <1/2
cho C=1/10+1/11+1/12+...+1/99+1/100
chứng tỏ C >1
a, Ta có: \(A=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{50}=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}\right)\)
Nhận xét: \(\frac{1}{11}+\frac{1}{12}+....+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{20}{30}=\frac{2}{3}\)
\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=\frac{20}{60}=\frac{1}{3}\)
\(\Rightarrow A>\frac{2}{3}+\frac{1}{3}=1>\frac{1}{2}\)
Vậy A > 1/2
b, Ta có: \(\frac{1}{50}>\frac{1}{100};\frac{1}{51}>\frac{1}{100};........;\frac{1}{99}>\frac{1}{100}\)
\(\Rightarrow B>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{50}{100}=\frac{1}{2}\)
Vậy B > 1/2
c, Ta có: \(C=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)\)
Nhận xét: \(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{90}{100}=\frac{9}{10}\)
\(\Rightarrow C>\frac{1}{10}+\frac{9}{10}=\frac{10}{10}=1\)
Vậy C > 1
CMR
1/50 +1/51 +1/52+...+1/98+1/99>1/2
So sánh
1/11+1/12+...+1/19+1/20 với 1/2
Bài này thầy Chung dạy rồi mà
\(\frac{99^1}{1}+\frac{99^2}{1}+\frac{99^3}{1}+...+\frac{99^{100}}{1}\)so sánh với 1001 vạn