Question

a) So sánh: A=\(\frac{100^{2009}+1}{100^{2008}+1}\)và B=\(\frac{100^{2010}+1}{100^{2009}+1}\)

b) Chứng minh rằng: \(\frac{1}{6}\)<\(\frac{1}{5^2}\)+\(\frac{1}{6^2}\)+\(\frac{1}{7^2}\)+...+\(\frac{1}{100^2}\)<\(\frac{1}{4}\)

Answer 1

\(a)\) 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 ~