So sánh 3^21 và 2^31

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

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

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

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

Đặt \(C=\dfrac{5^9}{1+5+5^2+...+5^9}\) ; \(D=\dfrac{3^9}{1+3+3^2+...+3^9}\)

Ta lại có: \(\dfrac{1}{C}=\dfrac{1+5+5^2+...+5^9}{5^9}\)

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

\(=\dfrac{1}{5^9}+\dfrac{1}{5^8}+\dfrac{1}{5^7}+...+\dfrac{1}{5}\)

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

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

\(=\dfrac{1}{3^9}+\dfrac{1}{3^8}+\dfrac{1}{3^7}+...+\dfrac{1}{3}\)

Vì \(\dfrac{1}{5^9}>\dfrac{1}{3^9};\dfrac{1}{5^8}>\dfrac{1}{3^8};....;\dfrac{1}{5}>\dfrac{1}{3}\)

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

\(\Rightarrow\dfrac{1}{C}< \dfrac{1}{D}\Rightarrow C>D\Rightarrow1+C>1+D\)

Mà \(1+C=A;1+D=B\) \(\Rightarrow A>B\)

Vậy A>B

b, A = 1 + 5 + 5 2 + . . . + 5 8 + 5 9 1 + 5 + 5 2 + . . . + 5 8 = 1 + 5 + 5 2 + . . . + 5 8 1 + 5 + 5 2 + . . . + 5 8 + 5 9 1 + 5 + 5 2 + . . . + 5 8 = 1 + 5 9 1 + 5 + 5 2 + . . . + 5 8 B = 1 + 3 + 3 2 + . . . + 3 9 1 + 3 + 3 2 + . . . + 3 8 = 1 + 3 + 3 2 + . . . + 3 8 1 + 3 + 3 2 + . . . + 3 8 + 3 9 1 + 3 + 3 2 + . . . + 3 8 Đặt C = 5 9 1 + 5 + 5 2 + . . . + 5 9 ; D = 3 9 1 + 3 + 3 2 + . . . + 3 9 Ta lại có: 1 C = 1 + 5 + 5 2 + . . . + 5 9 5 9 = 1 5 9 + 5 5 9 + 5 2 5 9 + . . . + 5 9 5 9 = 1 5 9 + 1 5 8 + 1 5 7 + . . . + 1 5 1 D = 1 + 3 + 3 2 + . . . + 3 9 3 9 = 1 3 9 + 3 3 9 + 3 2 3 9 + . . . + 3 9 3 9 = 1 3 9 + 1 3 8 + 1 3 7 + . . . + 1 3 Vì 1 5 9 > 1 3 9 ; 1 5 8 > 1 3 8 ; . . . . ; 1 5 > 1 3 ⇒ 1 5 9 + 1 5 8 + . . . . + 1 5 > 1 3 9 + 1 3 8 + . . . + 1 3 ⇒ 1 C < 1 D ⇒ C > D ⇒ 1 + C > 1 + D Mà 1 + C = A ; 1 + D = B ⇒ A > B Vậy A>B