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1+1+11+11+1+1+1+1+1+1+1+1+1+111+1+1+1+1+1+1+1+1+1+1111+1+1+1+11+1+1+1+1+1+1+1+1+1+1=1288
T= \(\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}.\dfrac{8}{9}.\dfrac{10}{11}.\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}.\dfrac{7}{8}.\dfrac{9}{10}\)
T=\(\dfrac{2.4.6.8.10.1.3.5.7.9}{3.5.7.9.11.2.4.6.8.10}\)
T=\(\dfrac{1.1.1.1.1.1.1.1.1.1}{1.1.1.1.11.1.1.1.1.1}\)
T=\(\dfrac{1}{11}\)
T=(1-1/2).(1-1/3).(1-1/4)....(1-1/10)
T=1/2.2/3.3/4.4/5....9/10
T=1.9/10
T=9/10
BÀI TOÁN DỄ NHẤT ĐẤY
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
1+1-1+1-1+1-1+1-1+1-2+1-1+1+1+1+1-1+1-1+1-1+1-1+1-1+1-1+1-1+1-1+1-1+1+1-11-1-1+1-1+1-1=1
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