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a: \(B=\left(-\dfrac{1}{5}-\dfrac{5}{7}+\dfrac{-3}{35}\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{2}\right)+\dfrac{1}{41}\)
\(=\dfrac{-7-25-3}{35}+\dfrac{3+2+1}{6}+\dfrac{1}{41}=\dfrac{42}{41}-1=\dfrac{1}{41}\)
a) \(\frac{1}{n}\) - \(\frac{1}{n+1}\) = \(\frac{n+1}{n\left(n+1\right)}\) - \(\frac{n}{n\left(n+1\right)}\) = \(\frac{1}{n\left(n+1\right)}\) = \(\frac{1}{n}\) . \(\frac{1}{n+1}\) =>đpcm
b) A= \(\frac{1}{2}\) - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\)+...+\(\frac{1}{8}\) - \(\frac{1}{9}\) +\(\frac{1}{9}\)
= \(\frac{1}{2}\) + \(\frac{1}{9}\)= \(\frac{11}{18}\)
a) Ta có: \(\frac{-9}{80}=\frac{\left(-9\right)x4}{80x4}=\frac{-36}{320}\) và \(\frac{17}{320}\)
b) Ta có: \(\frac{-7}{10}=\frac{\left(-7\right)x33}{10x33}=\frac{-231}{330}\) và \(\frac{1}{33}=\frac{1x10}{33x10}=\frac{10}{330}\)
c) Ta có:
\(\frac{-5}{14}=\frac{\left(-5\right)x10}{14x10}=\frac{-50}{140}\)
\(\frac{3}{20}=\frac{3x7}{20x7}=\frac{21}{140}\)
\(\frac{9}{70}=\frac{9x2}{70x2}=\frac{18}{140}\)
d) Ta có:
\(\frac{10}{42}=\frac{10x22}{42x22}=\frac{220}{924}\)
\(\frac{-3}{28}=\frac{\left(-3\right)x33}{28x33}=\frac{-99}{924}\)
\(\frac{-55}{132}=\frac{\left(-55\right)x7}{132x7}=\frac{-385}{924}\)
B=\(\frac{2011^{10}-1}{2011^{10}-3}\) <1 => \(\frac{2011^{10}-1}{2011^{10}-3}\) < \(\frac{2011^{10}-1+2}{2011^{10}-3+2}\) = \(\frac{2011^{10}+1}{2011^{10}-1}\) = A
=> B<A
M=1+1/2^2+1/3^2+1/4^2+...+1/10^2>1+1/2*3+1/3*4+1/4^5+...+1/10*11
M>1+1/2-1/3+1/4-1/4+1/5-...-1/11
M>1+1/2-1/11
M>1+9/22
M>31/22
vì 31/22>4/3 nên M>4/3
Ta thấy:\(\frac{5^{11}+1}{5^{10}+1}\)>1 nên theo quy tắc : \(\frac{a}{m}\)>1 thì \(\frac{a}{m}\)>\(\frac{a+m}{b+m}\) ta có:
B=\(\frac{5^{11}+1}{5^{10}+1}\)>\(\frac{5^{11}+1+4}{5^{10}+1+4}\)>\(\frac{5^{11}+5}{5^{10}+5}\)=\(\frac{5\left(5^{10}+1\right)}{5\left(5^9+1\right)}\)=A
Vậy B>A
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