Question

Cho \(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}\)

\(B=\frac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}\)

Chứng minh A > B  

help me please

Answer 1

\(A=1+\frac{5^9}{1+5+..+5^8}\)

      \(=1+\frac{1}{\frac{1}{5^9}+\frac{1}{5^8}+...+\frac{1}{5}}\)

Tương tự:

  \(B=1+\frac{1}{\frac{1}{3^9}+\frac{1}{3^8}+...+\frac{1}{3}}\)

Vì \(\frac{1}{5}< \frac{1}{3}\) , \(\frac{1}{5^2}< \frac{1}{3^2}\), . . .

nên: \(\frac{1}{\frac{1}{5^9}+\frac{1}{5^8}+...+\frac{1}{5}}>\frac{1}{\frac{1}{3^9}+\frac{1}{3^8}+...+\frac{1}{3}}\)

=> A > B

Vậy đề bạn cho chứng minh A < B là sai nhé.

Answer 2

Ta có:\(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}\)

=>\(A=\frac{\left(1+5+5^2+...+5^8\right)}{\left(1+5+5^2+...+5^8\right)}+\frac{5^9}{1+5+5^2+...+5^8}\)

=>\(A=1+\frac{5^9}{1+5+5^2+...+5^8}\)

Ta có:\(B=\frac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}\)

=>\(B=\frac{1+3+3^2+...+3^8}{1+3+3^2+...+3^8}+\frac{3^9}{1+3+3^2+...+3^8}\)

=>\(B=1+\frac{3^9}{1+3+3^2+...+3^8}\)

vì:\(1+3+3^2+...+3^8< 1+5+5^2+...+5^8\)

Nên A<B(đpcm).