So sánh A và B biết A=(1+5+5^2+...+5^9)/(1+5+5^2+...+5^8) và B=...

Theo bài ra, ta có:

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

= \(\dfrac{1+5+5^2+...+5^8}{1+5+5^2+...+5^8}\)+ \(\dfrac{5^9}{1+5+5^2+...+5^8}\)

= 1 + \(\dfrac{1}{\dfrac{1+5+5^2+...+5^8}{5^9}}\)

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

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

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

Nhận xét:

+) \(\dfrac{1+5+5^2+...+5^8}{5^9}\) = \(\dfrac{1}{5^9}\) + \(\dfrac{1}{5^8}\) + ... + \(\dfrac{1}{5^{ }}\)

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

Có: \(\dfrac{1}{5^9}\) < \(\dfrac{1}{3^9}\) ; \(\dfrac{1}{5^8}\) < \(\dfrac{1}{3^8}\) ; ... ; \(\dfrac{1}{5^{ }}\) < \(\dfrac{1}{3}\)

⇒ \(\dfrac{1+5+5^2+...+5^8}{5^9}\) < \(\dfrac{1+3+3^2+...+3^8}{3^9}\)

⇒ \(\dfrac{1}{\dfrac{1+5+5^2+...+5^8}{5^9}}\) > \(\dfrac{1}{\dfrac{1+3+3^2+...+3^8}{3^9}}\)

⇒ A > B

Vậy A > B.

Giải:

a) Biến đổi tử:

Đặt:

\(C=1+5+5^2+5^3+...+5^9\)

\(\Leftrightarrow5C=5+5^2+5^3+5^4...+5^{10}\)

\(\Leftrightarrow5C-C=5^{10}-1\)

\(\Leftrightarrow4C=5^{10}-1\)

\(\Leftrightarrow C=\dfrac{5^{10}-1}{4}\)

Tương tự ta có mẫu là:

\(\dfrac{5^9-1}{4}\)

Đặt vào A, được:

\(A=\dfrac{1+5+5^2+5^3+...+5^9}{1+5+5^2+5^3+...+5^8}\)

\(\Leftrightarrow A=\dfrac{\dfrac{5^{10}-1}{4}}{\dfrac{5^9-1}{4}}\)

\(\Leftrightarrow A=\dfrac{5^{10}-1}{5^9-1}\)

Vậy ...

b) Tương tự câu a, ta được:

\(B=\dfrac{\dfrac{3^{10}-1}{2}}{\dfrac{3^9-1}{2}}\)

\(\Leftrightarrow B=\dfrac{3^{10}-1}{3^9-1}\)

Vậy ...