Ta có : \(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{1599}{1600}\)

\(=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{6}\right)...\left(1-\frac{1}{1600}\right)\)

Đặt \(B=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{1600}{1601}\)

\(=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right)...\left(1-\frac{1}{1601}\right)\)

Vì \(\frac{1}{2}>\frac{1}{3};\frac{1}{4}>\frac{1}{5};\frac{1}{6}>\frac{1}{7};...;\frac{1}{1600}>\frac{1}{1601}\)

\(\Rightarrow1-\frac{1}{2}< 1-\frac{1}{3};1-\frac{1}{4}< 1-\frac{1}{5};1-\frac{1}{6}< 1-\frac{1}{7};...;1-\frac{1}{1600}< 1-\frac{1}{1601}\)

\(\Rightarrow A< B\)

hay A<\(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{1600}{1601}\)

Vậy A<\(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{1600}{1601}\).

Ta luôn có: 

\(\frac{1}{2}< \frac{2}{3}\)

\(\frac{3}{4}< \frac{4}{5}\)

\(\frac{5}{7}< \frac{6}{7}\)

\(........\)

\(\frac{1599}{1600}< \frac{1600}{1601}\)

Từ trên: \(\Rightarrow A=\frac{1}{2}.\frac{3}{4}....\frac{1599}{1600}\left(1\right)\)

\(\Rightarrow\frac{1}{2}.\frac{3}{4}...\frac{1599}{1600}< \frac{2}{3}.\frac{4}{5}....\frac{1600}{1601}\left(2\right)\)

Từ: \(\left(1\right)\left(2\right)\Rightarrow A< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{1600}{1601}\left(đpcm\right)\)

bạn ơi trả lời họ mình với

là sao

vào trả lời hộ

Ta có

 \(\frac{1}{2}< \frac{2}{3},\frac{3}{4}< \frac{4}{5},...,\frac{1599}{1600}< \frac{1600}{1601}\)

Do đó ta có

A=\(\frac{1}{2}\times\frac{3}{4}\times...\times\frac{1599}{1600}< \frac{2}{3}\times\frac{4}{5}\times...\times\frac{1600}{1601}\)

#Châu's ngốc

Đề thiếu. Bạn xem lại đề.

\(\dfrac{n^2-1}{n^2}=1-\dfrac{1}{n^2}>1-\dfrac{1}{\left(n-1\right)n}\)

Từ đó ta có:

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

\(\Rightarrow A>49-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)\)

\(\Rightarrow A>49-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)

\(\Rightarrow A>49-\left(1-\dfrac{1}{50}\right)=48+\dfrac{1}{50}>48\)

\(A=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{2499}{2500}\\ A=\left(1+1+1+...+1\right)-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}\right)\\ A=49-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}\right)\)

Có \(\dfrac{1}{4}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\\ \dfrac{1}{9}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\\ \dfrac{1}{16}=\dfrac{1}{4.4}< \dfrac{1}{3.4}\\ ...\\ \dfrac{1}{2500}=\dfrac{1}{50.50}< \dfrac{1}{49.50}\)

\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\\ \Rightarrow\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}< 1-\dfrac{1}{50}< 1\\ \Rightarrow\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}< 1\)

\(\Rightarrow A=49-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}\right)>49-1\\ \Rightarrow A>48\)