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Ta có: \(\left(\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{9}\right)>\dfrac{1}{9}.6=\dfrac{6}{9}>\dfrac{1}{2}\) (1)
\(\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{19}\right)>\dfrac{1}{19}.10=\dfrac{10}{19}>\dfrac{1}{2}\) (2)
\(\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{19}>\left(1\right)+\left(2\right)\)
\(\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{19}>1\left(đpcm\right)\)
\(\frac{3}{5}-\frac{-7}{10}-\frac{13}{-20}=\frac{3}{5}+\frac{7}{10}+\frac{13}{20}=\frac{12}{20}+\frac{14}{20}+\frac{13}{20}=\frac{39}{20}\)
\(\frac{1}{2}+\frac{-1}{3}+\frac{1}{4}-\frac{-1}{6}=\frac{1}{2}+\frac{-1}{3}+\frac{1}{4}+\frac{1}{6}=\frac{6+(-4)+3+2}{12}=\frac{7}{12}\)
\(\frac{9}{4}.\frac{8}{27}.\frac{5}{7}=\frac{9.8.5}{4.27.7}=\frac{1.2.5}{1.3.7}=\frac{10}{21}\)
\(\frac{2}{5}.(\frac{2}{3}-\frac{1}{4})+\frac{1}{2}=\frac{2}{5}.(\frac{8}{12}-\frac{3}{12})+\frac{1}{2}=\frac{2}{5}.\frac{5}{12}+\frac{1}{2}=\frac{1}{6}+\frac{1}{2}=\frac{1}{6}+\frac{3}{6}=\frac{4}{6}=\frac{2}{3}\)
\((\frac{1}{3}-\frac{1}{6}):(\frac{1}{3}+\frac{1}{6})=(\frac{2}{6}-\frac{1}{6}):(\frac{2}{6}+\frac{1}{6})=\frac{1}{6}:\frac{3}{6}=\frac{1}{6}.\frac{6}{3}=\frac{1.6}{6.3}=\frac{1.1}{1.3}=\frac{1}{3}\)
Hok tốt
Ta có:
1 = \(\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+............+\frac{1}{10}\)(10 phân số \(\frac{1}{10}\))
Mà \(\frac{1}{2}>\frac{1}{10};\frac{2}{3}>\frac{1}{10};............;\frac{9}{10}>10\)
\(\Rightarrow M>1\)
Vậy M > 1
có:
(1994-1)+1=1994
Tổng là:
1994x(1994+1):2=1989015
Đáp số:1989015
Ta có:\(\frac{1}{2}>\frac{1}{8};\frac{1}{3}>\frac{1}{8};...;\frac{1}{6}>\frac{1}{8};\frac{1}{7}+\frac{1}{8}+\frac{1}{9}>\frac{3}{8}\)
\(\Rightarrow\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{3}{8}\)
\(=\frac{8}{8}=1\)
Vậy\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}>1\)