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a, Ta có:
\(\frac{-3}{4}=\frac{-15}{20}< \frac{-7}{20}\Rightarrow\frac{-3}{4}< \frac{-7}{20}\)
b,Ta có:\(\frac{-7}{8}< 1< \frac{30}{-42}\Rightarrow\frac{-7}{8}< \frac{30}{-42}\)
Thank:)
Đặt \(B=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot....\cdot\frac{10000}{10001}\)
\(\Rightarrow A< B\)
\(\Rightarrow A^2< AB\)
\(\Rightarrow A^2< \left(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot....\cdot\frac{9999}{10000}\right)\left(\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot....\cdot\frac{10000}{10001}\right)\)
\(=\frac{1}{10001}< \frac{1}{10000}=0.0001\)
\(\Rightarrow A^2< 0.0001\)
\(\Rightarrow A< 0.1\)
a) A = \(\frac{101}{19}.\) \(\frac{61}{218}-\frac{101}{218}.\frac{42}{19}+\frac{117}{218}\)
= \(\frac{101}{218}.\frac{61}{19}-\frac{101}{218}.\frac{42}{19}+\frac{117}{218}\)
=\(\frac{101}{218}.\left(\frac{61}{19}-\frac{42}{19}\right)+\frac{117}{218}\)
=\(\frac{101}{218}.\frac{19}{19}+\frac{117}{218}\)
=\(\frac{101}{218}.1+\frac{117}{218}\)
=\(\frac{101}{218}+\frac{117}{218}\)
=\(\frac{218}{218}\)\(=1\)
b) B = \(\left(\frac{5}{2011^2}+\frac{7}{2012^2}-\frac{9}{2013^2}\right).\left(\frac{4}{5}-\frac{3}{4}-\frac{1}{20}\right)\)
= \(\left(\frac{5}{2011^2}+\frac{7}{2012^2}-\frac{9}{2013^2}\right)\)\(.\left(\frac{1}{20}-\frac{1}{20}\right)\)
= \(\left(\frac{5}{2011^2}+\frac{7}{2012^2}-\frac{9}{2013^2}\right).0\)
= \(0\)
Ta có :\(C=\frac{20^{10}+1}{20^{10}-1}\)
=> \(C-1=\frac{20^{10}+1-\left(20^{10}-1\right)}{20^{10}-1}=\frac{2}{20^{10}-1}\)
Lại có D = \(\frac{20^{10}-1}{20^{10}-3}\)
=> D - 1 = \(\frac{20^{10}-1-\left(20^{10}-3\right)}{20^{10}-3}=\frac{2}{20^{10}-3}\)
Vì \(20^{10}-1>20^{10}-3\Rightarrow\frac{2}{20^{10}-1}< \frac{2}{2^{10}-3}\Rightarrow C-1< D-1\Rightarrow C< D\)
Có : \(C=\frac{20^{10}+1}{20^{10}-1}\)
< = > \(C-1=\frac{20^{10}+1-\left(20^{10}-1\right)=\frac{2}{20^{10}-1}}{20^{10}-1}\)
có D \(\frac{20^{10}-1}{20^{10}-3}\)
=> D - 1 = \(\frac{20^{10}-1\left(20^{10}-3\right)}{20^{10}-3}=\frac{2}{20^{10}-3}\)