\(B=\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{37.39}=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{39}\)

\(B=\dfrac{1}{3}-\dfrac{1}{39}=\dfrac{12}{39}-\dfrac{1}{39}=\dfrac{11}{39}\)

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(A=\dfrac{1}{1}-\dfrac{1}{50}=\dfrac{50}{50}-\dfrac{1}{50}=\dfrac{49}{50}\)

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

a, Ta có: \(3^{21}>3^{20}\left(1\right)\)

\(2^{31}>2^{30}\)(2)

Mà \(\left\{{}\begin{matrix}3^{20}=3^{2.10}=\left(3^2\right)^{10}=9^{10}\\2^{30}=2^{3.10}=\left(2^3\right)^{10}=8^{10}\end{matrix}\right.\)

Do \(9>8\Rightarrow9^{10}>8^{10}\Rightarrow3^{20}>2^{30}\left(3\right)\)

Từ (1);(2) và (3) ta suy ra \(3^{21}>2^{31}\)

Ta có: \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{63}>\dfrac{1}{32}+\dfrac{1}{32}+..+\dfrac{1}{32}\left(có\right)62sốhạng\)

\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{63}>\dfrac{1}{32}.63=\dfrac{63}{32}=2\)

\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{63}>2\)(đây là điều cần chứng tỏ)

có trên online math ấy. Vào mạng đánh rồi nhấn enter sau đó có dòng đầu tiên nháy chuột vào đó là có