Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
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
\(B=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)...\left(1-\dfrac{1}{81}\right)\left(1-\dfrac{1}{100}\right)\)
\(=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}...\dfrac{99}{100}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}...\dfrac{9.11}{10.10}=\left(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{9}{10}\right).\left(\dfrac{3}{2}.\dfrac{4}{3}...\dfrac{11}{10}\right)=\dfrac{1}{10}.\dfrac{11}{2}=\dfrac{11}{20}>\dfrac{11}{21}\)
\(B=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)...\left(1-\dfrac{1}{9}\right)\left(1+\dfrac{1}{9}\right)\left(1-\dfrac{1}{10}\right)\left(1+\dfrac{1}{10}\right)\\ B=\left(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{8}{9}\cdot\dfrac{9}{10}\right)\left(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{10}{9}\cdot\dfrac{11}{10}\right)\\ B=\dfrac{1}{10}\cdot\dfrac{11}{2}=\dfrac{11}{20}>\dfrac{11}{21}\)
a: \(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{10}-1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-9}{10}\)
\(=-\dfrac{1}{10}\)
9<10
=>1/9>1/10
=>\(-\dfrac{1}{9}< -\dfrac{1}{10}\)
=>\(A>-\dfrac{1}{9}\)
b: \(B=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{10}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{10}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-9}{10}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{11}{10}\)
\(=\dfrac{-1}{10}\cdot\dfrac{11}{2}=\dfrac{-11}{20}\)
20<21
=>\(\dfrac{11}{20}>\dfrac{11}{21}\)
=>\(-\dfrac{11}{20}< -\dfrac{11}{21}\)
=>\(B< -\dfrac{11}{21}\)
Ta có:
\(A=\left(\dfrac{1}{2}-1\right)\cdot\left(\dfrac{1}{3}-1\right)\cdot\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{10}-1\right)\)
\(A=-\dfrac{1}{2}\cdot-\dfrac{2}{3}-\dfrac{3}{4}\cdot...\cdot-\dfrac{9}{10}\)
\(A=\dfrac{-1\cdot-2\cdot-3\cdot...\cdot-9}{2\cdot3\cdot4\cdot...\cdot10}\)
\(A=-\dfrac{1}{10}\)
Mà: \(10>9\)
\(\Rightarrow\dfrac{1}{10}< \dfrac{1}{9}\)
\(\Rightarrow-\dfrac{1}{10}>-\dfrac{1}{9}\)
\(\Rightarrow A>-\dfrac{1}{9}\)
\(A=\left(\dfrac{1}{4}-1\right).\left(\dfrac{1}{9}-1\right)....\left(\dfrac{1}{100}-1\right).\)
\(\Rightarrow A=\left(-\dfrac{3}{4}\right).\left(-\dfrac{8}{9}\right)....\left(-\dfrac{99}{100}\right)\)
mà A có 9 dấu - \(\left(4;9;16;25;36;49;64;81;100\right)\)
\(\Rightarrow0>A=\left(-\dfrac{3}{4}\right).\left(-\dfrac{8}{9}\right)....\left(-\dfrac{99}{100}\right)=-\dfrac{1}{2}\)
Ta lại có \(\left\{{}\begin{matrix}\dfrac{1}{2}=\dfrac{21}{42}\\\dfrac{11}{21}=\dfrac{22}{42}\end{matrix}\right.\) \(\Rightarrow\dfrac{1}{2}< \dfrac{11}{21}\Rightarrow-\dfrac{1}{2}>-\dfrac{11}{21}\)
\(\Rightarrow A>-\dfrac{11}{21}\)
\(A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)...\left(\dfrac{1}{100}-1\right)\)
\(A=\left(-\dfrac{2^2-1}{2^2}\right)\left(-\dfrac{3^2-1}{3^2}\right)...\left(-\dfrac{10^2-1}{10^2}\right)\)
\(A=\left[-\dfrac{1\cdot3}{2\cdot2}\right]\left[-\dfrac{2\cdot4}{3\cdot3}\right]...\left[-\dfrac{9\cdot11}{10\cdot10}\right]\)
Dễ thấy A có 9 thừa số, suy ra
\(A=-\dfrac{1\cdot3\cdot2\cdot4\cdot...\cdot9\cdot11}{2\cdot2\cdot3\cdot3\cdot...\cdot10.10}=-\dfrac{1\cdot11}{2\cdot10}=\dfrac{-11}{20}\)
Vì 20 < 21 nên \(\dfrac{11}{20}>\dfrac{11}{21}\), suy ra \(\dfrac{-11}{20}< \dfrac{-11}{21}\)
Vậy \(A< \dfrac{-11}{21}\)
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 \(\left(\frac{9}{11}-0,81\right)^{2005}< \frac{1}{10^{4010}}\)
Đặt \(\dfrac{1}{5}+\dfrac{2013}{2014}+\dfrac{2015}{2016}=B;\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}=C\)
\(A=\left(B+1\right)\cdot C-B\cdot\left(C+1\right)\)
\(=BC+C-BC-B\)
=C-B
\(=\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}-\dfrac{1}{5}-\dfrac{2013}{2014}-\dfrac{2015}{2016}=-\dfrac{1}{10}\)
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 \(M< N\)