Question

B1: 

Cho A = \(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\). Hãy chứng tỏ rằng A > 1

B2:

So sánh: \(B=\) \(\frac{20^{10}+1}{20^{10}-1}\)và \(C=\frac{20^{10}-1}{20^{10}-3}\)

Answer 1

B2 :P Ta có : \(B=\frac{2^{10}+1}{2^{10}-1}=1+\frac{2}{2^{10}-1}\)

      \(C=\frac{2^{10}-1}{2^{10}-3}=1+\frac{2}{2^{10}-3}\)

Nên : B > C

Answer 2

Nhầm C > B