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Ta có:
\(M=\left(\dfrac{9}{11}-0,81\right)^{2014}\)
\(=\left(\dfrac{9}{1100}\right)^{2014}\)
\(=\left(\dfrac{9}{11}.\dfrac{1}{100}\right)^{2014}\)
\(=\dfrac{9^{2014}}{11^{2014}}.\dfrac{1}{100^{2.2014}}\)
\(=\dfrac{9^{2014}}{11^{2014}}.\dfrac{1}{10^{2048}}\)
Vì: \(\dfrac{1}{10^{2048}}=\dfrac{1}{10^{2048}}\)
Nên: \(\dfrac{9^{2014}}{11^{2014}}.\dfrac{1}{10^{2048}}>\dfrac{1}{10^{2048}}\)
Hay: M>N
Vậy M>N
\(\dfrac{9}{11}-0,81=\dfrac{9}{11}-\dfrac{81}{100}=\dfrac{81}{99}-\dfrac{81}{100}< \dfrac{81+1}{99+1}-\dfrac{81}{100}\)
\(=\dfrac{82}{100}-\dfrac{81}{100}=\dfrac{1}{100}\)
\(\Rightarrow\left(\dfrac{9}{11}-0,81\right)^{2014}< \left(\dfrac{1}{100}\right)^{2014}=\dfrac{1}{10^{4028}}\)
\(M< N\)
Vậy \(M< N\)
a: =>4x-6-9=5-3x-3
=>4x-15=-3x+2
=>7x=17
hay x=17/7
b: \(\Leftrightarrow\dfrac{2}{3x}-\dfrac{1}{4}=\dfrac{4}{5}-\dfrac{7}{x}+2\)
=>2/3x+21/3x=4/5+2+1/4=61/20
=>23/3x=61/20
=>3x=23:61/20=460/61
hay x=460/183
\(B=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)...\left(1-\dfrac{1}{81}\right)\left(1-\dfrac{1}{100}\right)\)
\(=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}...\dfrac{99}{100}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}...\dfrac{9.11}{10.10}=\left(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{9}{10}\right).\left(\dfrac{3}{2}.\dfrac{4}{3}...\dfrac{11}{10}\right)=\dfrac{1}{10}.\dfrac{11}{2}=\dfrac{11}{20}>\dfrac{11}{21}\)
\(B=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)...\left(1-\dfrac{1}{9}\right)\left(1+\dfrac{1}{9}\right)\left(1-\dfrac{1}{10}\right)\left(1+\dfrac{1}{10}\right)\\ B=\left(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{8}{9}\cdot\dfrac{9}{10}\right)\left(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{10}{9}\cdot\dfrac{11}{10}\right)\\ B=\dfrac{1}{10}\cdot\dfrac{11}{2}=\dfrac{11}{20}>\dfrac{11}{21}\)
Ta có:\(\left(\frac{9}{11}-0,81\right)^{2005}\)=\(\left(\frac{9}{11}-\frac{81}{100}\right)^{2005}=\left(\frac{9}{1100}\right)^{2005}< \left(\frac{10}{1100}\right)^{2005}=\left(\frac{1}{110}\right)^{2005}\)
Mà \(\left(\frac{1}{110}\right)^{2005}< \left(\frac{1}{100}\right)^{2005}=\left[\left(\frac{1}{10}\right)^2\right]^{2005}=\left(\frac{1}{10}\right)^{4010}=\frac{1}{10^{4010}}\)
Vậy \(\left(\frac{9}{11}-0,81\right)^{2005}< \frac{1}{10^{4010}}\)
Ta có:
\(\left(\dfrac{1}{10}\right)^{15}=\left(\left(\dfrac{1}{10}\right)^3\right)^5=\left(\dfrac{1}{1000}\right)^5\)
\(\left(\dfrac{3}{10}\right)^{20}=\left(\left(\dfrac{3}{10}\right)^4\right)^5=\left(\dfrac{81}{10000}\right)^5\)
Ta có: \(\left(\dfrac{1}{10}\right)^{15}=\left(\dfrac{1}{10}^3\right)^5=\left(\dfrac{1}{1000}\right)^5\)
\(\left(\dfrac{3}{10}\right)^{20}=\left(\dfrac{3}{10}^4\right)^5=\left(\dfrac{3}{10000}\right)^5\)
Vì \(\dfrac{1}{1000}>\dfrac{3}{10000}\) nên \(\left(\dfrac{1}{10}\right)^{15}>\left(\dfrac{3}{10}\right)^{20}\)
Ta có: \(\left(\dfrac{9}{11}-0,81\right)^{2014}\)
\(=\left(\dfrac{9}{11}-\dfrac{81}{100}\right)^{2014}\)
\(=\left(\dfrac{900}{1100}-\dfrac{891}{1100}\right)^{2014}\)
\(=\left(\dfrac{9}{1100}\right)^{2014}\)
\(=\left(\dfrac{9}{11}.\dfrac{1}{100}\right)^{2014}\)
\(=\left(\dfrac{9}{11}\right)^{2014}.\dfrac{1}{10^{4028}}\)
\(=\left(\dfrac{9}{11}\right)^{2014}.N\)
Vì \(\left(\dfrac{9}{11}\right)^{2014}< 1\) nên M < N