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\(a)\) Ta có :
\(\frac{1}{100}A=\frac{100^{2009}+1}{100^{2009}+100}=\frac{100^{2009}+100}{100^{2009}+100}-\frac{99}{100^{2009}+100}=1-\frac{99}{100^{2009}+100}\)
\(\frac{1}{100}B=\frac{100^{2010}+1}{100^{2010}+100}=\frac{100^{2010}+100}{100^{2010}+100}-\frac{99}{100^{2010}+100}=1-\frac{99}{100^{2010}+100}\)
Vì \(\frac{99}{100^{2009}+100}>\frac{99}{100^{2010}+100}\) nên \(1-\frac{99}{100^{2009}+100}< 1-\frac{99}{100^{2010}+100}\)
Do đó :
\(\frac{1}{100}A< \frac{1}{100}B\)\(\Rightarrow\)\(A< B\)
Vậy \(A< B\)
Chúc bạn học tốt ~
\(A=\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
\(A>\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{100.101}\)
\(A>\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+....+\frac{1}{100}-\frac{1}{101}\)
\(A>\frac{1}{5}-\frac{1}{101}=\frac{100}{505}>\frac{100}{600}=\frac{1}{6}\)
Tương tự
\(A< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{99}-\frac{1}{100}\)
\(A< \frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)
a) Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}\) ; \(\frac{1}{3^2}< \frac{1}{2.3}\) ; \(\frac{1}{4^2}< \frac{1}{3.4}\) ; ... ; \(\frac{1}{2010^2}< \frac{1}{2009.2010}\)
=> \(Vt< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(=1-\frac{1}{2010}< 1\)
Ta xét A= \(\frac{1}{5^2}+\frac{1}{6^2}+..+\frac{1}{100^2}\)
\(\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}>\frac{1}{5.6}+\frac{1}{6.7}...+\frac{1}{100.101}\)
=> \(A>\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)
=> \(A>\frac{1}{5}-\frac{1}{101}\)
=> \(A>\frac{96}{505}>\frac{96}{576}=\frac{1}{4}\)
Ta có : \(A< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
=> \(A< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
=> \(A< \frac{1}{4}-\frac{1}{100}\)
=> \(A< \frac{6}{25}< \frac{6}{24}=\frac{1}{4}\)
\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+.....+\frac{1}{100^2}< \frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{50^2}\right)\)
\(\Leftrightarrow\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3}............\)
\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+........+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+.......+\frac{1}{49.50}\)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+.....+\frac{1}{49}-\frac{1}{50}\)
\(\Leftrightarrow1-\frac{1}{50}\)
\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{50^2}\Rightarrow1+\frac{1}{2^2}+....+\frac{1}{50^2}< 1+1=2\)
\(\Leftrightarrow\frac{1}{2^2}.\left(1+\frac{1}{2^2}+....+\frac{1}{50^2}\right)< \frac{1}{2^2}.2=\frac{1}{2}\)
\(\LeftrightarrowĐPCM\)