cho= A= 1+5+5²+...+5⁹/1+5+5²+...+5⁸ Cho B =1+3+3²+..+3⁹/1+3+3²+3⁸ CMR A>B

$#shin$

Giải thích các bước giải:

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 ( dpcm )`