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a, Xét 2010 . 2010 = (2009+1).2010
= 2009.2010 +2010
= (2009.2010+2009)+1
= 2009.(2010+1)+1
= 2009.2011+1
>= 2009.2010
=> 2010/2009 > 2011/2010
Tk mk nha
a, \(\frac{2010}{2009}\)và \(\frac{2011}{2010}\)
Ta có:
2010.2010 = ( 2009 + 1 ) . 2010
= 2009 . 2010 + 2010
= ( 2009 . 2010 + 2019 ) + 1
= 2019 . ( 2010 + 1 ) + 1
= 2019 . 2011 + 1
\(\Rightarrow\)\(\frac{2010}{2009}>\frac{2011}{2010}\)
b, \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...........+\frac{1}{200}\)và 1
Ta có:
\(\frac{1}{101}< 1;\frac{1}{102}< 1;\frac{1}{103}< 1;........;\frac{1}{200}< 1\)
\(\Rightarrow\)\(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+.............+\frac{1}{200}< 1\)
\(a.\frac{1}{2^{300}}=\frac{1}{\left(2^3\right)^{100}}=\frac{1}{8^{100}}\)
\(\frac{1}{3^{200}}=\frac{1}{\left(3^2\right)^{100}}=\frac{1}{9^{100}}\)
\(\text{Vì }\frac{1}{8}>\frac{1}{9}\Rightarrow\frac{1}{\left(2^3\right)^{100}}>\frac{1}{\left(3^2\right)^{100}}\Rightarrow\frac{1}{2^{300}}>\frac{1}{3^{200}}\)
\(b.\frac{1}{5^{199}}:\text{Giữ nguyên}\)
\(\frac{1}{3^{200}}=\frac{1}{3^{199}\cdot3}\)
\(\frac{1}{5^{199}}< \frac{1}{3^{199}\cdot3}\Rightarrow\frac{1}{5^{199}}< \frac{1}{3^{200}}\)
2 bài dưới bn làm tương tự nhé
2. a) \(3^{200}=\left(3^2\right)^{100}=9^{100}\)
\(2^{300}=\left(2^3\right)^{100}=8^{100}\)
Vì \(9^{100}>8^{100}\Rightarrow3^{200}>2^{300}\)
b) \(71^{50}=\left(71^2\right)^{25}=5041^{25}\)
\(37^{75}=\left(3^3\right)^{25}=27^{25}\)
Vì \(5041^{25}>27^{25}\Rightarrow71^{50}>37^{75}\)
c) \(\frac{201201}{202202}=\frac{201201:1001}{202202:1001}=\frac{201}{202}\)
\(\frac{201201201}{202202202}=\frac{201201201:1001001}{202202202:1001001}=\frac{201}{202}\)
Vì \(\frac{201}{202}=\frac{201}{202}\Rightarrow\frac{201201}{202202}=\frac{201201201}{202202202}\)