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Ta có: \(\frac{1}{151}>\frac{1}{300};\frac{1}{152}>\frac{1}{300};\ldots;\frac{1}{300}=\frac{1}{300}\)
Do đó: \(\frac{1}{151}+\frac{1}{152}+\cdots+\frac{1}{300}>\frac{1}{300}+\frac{1}{300}+\cdots+\frac{1}{300}=\frac{150}{300}=\frac12\left(1\right)\)
Ta có: \(\frac{1}{301}>\frac{1}{450};\frac{1}{302}>\frac{1}{450};\ldots;\frac{1}{450}=\frac{1}{450}\)
Do đó; \(\frac{1}{301}+\frac{1}{302}+\cdots+\frac{1}{450}>\frac{1}{450}+\frac{1}{450}+\cdots+\frac{1}{450}=\frac{150}{450}=\frac13\) (2)
Từ (1),(2) suy ra \(\left(\frac{1}{151}+\frac{1}{152}+\cdots+\frac{1}{300}\right)+\left(\frac{1}{301}+\frac{1}{302}+\cdots+\frac{1}{450}\right)\) >1/2+1/3
=>\(\left(\frac{1}{151}+\frac{1}{152}+\cdots+\frac{1}{300}\right)+\left(\frac{1}{301}+\frac{1}{302}+\cdots+\frac{1}{450}\right)\) >5/6(ĐPCM)
\(\left(\frac{1}{3\cdot8}+\frac{1}{8\cdot13}+\cdots+\frac{1}{33\cdot38}\right)-x=\frac{31}{114}\)
=>\(\frac15\left(\frac{5}{3\cdot8}+\frac{5}{8\cdot13}+\cdots+\frac{5}{33\cdot38}\right)-x=\frac{31}{114}\)
=>\(\frac15\left(\frac13-\frac18+\frac18-\frac{1}{13}+\cdots+\frac{1}{33}-\frac{1}{38}\right)-x=\frac{31}{114}\)
=>\(\frac15\left(\frac13-\frac{1}{38}\right)-x=\frac{31}{114}\)
=>\(\frac15\cdot\frac{35}{3\cdot38}-x=\frac{31}{114}\)
=>\(\frac{7}{114}-x=\frac{31}{114}\)
=>\(x=\frac{7}{114}-\frac{31}{114}=-\frac{24}{114}=\frac{-4}{19}\)
- B=(1/2).(2/3).(3/4)....(2010/2011).(2011/2012)
B=(1.2.3....2011)/(2.3.4....2012)
B=1/2012
a) \(27^{64}:81^{20}=3^{192}:3^{80}=3^{112}\)
b) \(\left(\dfrac{1}{8}\right)^{20}:\left(\dfrac{1}{16}\right)^9=\left(\dfrac{1}{2}\right)^{60}:\left(\dfrac{1}{2}\right)^{36}=\left(\dfrac{1}{2}\right)^{24}\)
c) \(\dfrac{1}{3}:\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{5}{3}-\dfrac{1}{6}=\dfrac{10}{6}-\dfrac{1}{6}=\dfrac{9}{6}=\dfrac{3}{2}\)