ta có 1/2<2/3 ; 3/4<4/5;5/6<6/7;...;199/200<200/201

suy ra A^2=1/2^2*3/4^2*5/6^2*...*199/200^2<1/2*2/3*3/4*4/5*5/6*6/7*...*199/200/200/201

suy ra A^2<1/201(đpcm)

Ta có:

\(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};...;\frac{199}{200}< \frac{200}{201}\)

\(\Rightarrow\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{199}{200}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{200}{201}\)

\(\Rightarrow A< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{200}{201}\)

\(\Rightarrow A^2< \left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{200}{201}\right)\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{199}{200}\right)\)

\(\Rightarrow A^2< \frac{1}{201}\left(đpcm\right)\)

\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{199}{200}\)

\(\Rightarrow A< \frac{2}{3}.\frac{4}{5}\frac{6}{7}...\frac{200}{201}\)

\(\Rightarrow A.A< A.\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{200}{201}\right)\)

\(\Rightarrow A^2< \frac{1}{201}\)(làm phần trc như Sakuraba Laura nhá)

câu hỏi hay nhưng khó quá

Nguyễn Ngọc Sáng nói chí lí

1/201 > rồi cần hì cm và ss

Lời giải:

Ta có:

\(\text{VT}=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{200}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)

\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}=\text{VP}\)

Ta có đpcm.

Ta có:

\(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};...;\frac{199}{200}< \frac{200}{201}\)

\(\Rightarrow\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{199}{200}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{200}{201}\)

\(\Rightarrow C< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{200}{201}\)

\(\Rightarrow C^2< \left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{200}{201}\right).\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{199}{200}\right)\)

\(\Rightarrow C^2< \frac{1}{201}\left(dpcm\right)\)

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{199}-\frac{1}{200}\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{200}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{199}+\frac{1}{200}-1-\frac{1}{2}-\frac{1}{4}-....-\frac{1}{100}\)

\(=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)+\left(\frac{1}{101}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+...+\frac{1}{100}\right)\)

\(=\frac{1}{101}+\frac{1}{102}+.....+\frac{1}{199}+\frac{1}{200}\) (ĐPCM)

Ta có : 1 - 1/2 + 1/3 - 1/4 + ....- 1/200

= (1 + 1/3 + 1/5 + ....+ 1/199) - ( 1/2 + 1/4 + 1/6 + .... + 1/200)

= ( 1 + 1/3 +...+ 1/199) + (1/2 +1/4 + ...+ 1/200) - 2(1/2+1/4+...+ 1/200)

= (1+1/2+1/3+....+1/199 + 1/200) - (1 +1/2 +1/3 +....+1/100)

= 1/101 + 1/102+ 1/103 + .... + 1/200

chúc bạn học tốt!!!!!!!

Bạn tham khảo tại Câu hỏi của lê chí dũng - Chuyên mục hỏi đáp - Giúp tôi giải toán. - Học toán với OnlineMath

Chúc bạn học tốt!

Tks bạn nhé Nguyễn Thế Bảo

Ta có : \(1-\frac{1}{2}+\frac{1}{3}-...-\frac{1}{200}=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)

 \(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+...+\frac{1}{100}\right)\)

\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)\(\left(đpcm\right)\)