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Ta có \(k^2>k^2-1=\left(k+1\right)\left(k-1\right)\)
Áp dung vào bài toán ta được
\(A=\frac{1}{2}.\frac{3}{4}...\frac{199}{200}=\frac{1.3...199}{2.4...200}\)
\(\Rightarrow A^2=\frac{1^2.3^2...199^2}{2^2.4^2...200^2}< \frac{1^2.3^2...199^2}{1.3.3.5...199.201}=\frac{1^2.3^2...199^2}{1.3^2.5^2...199^2.201}=\frac{1}{201}\)
Vậy \(A^2< \frac{1}{201}\)
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:
\(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};...;\frac{199}{200}< \frac{200}{201}\)
\(\Rightarrow\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{199}{200}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{200}{201}\)
\(\Rightarrow A< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{200}{201}\)
\(\Rightarrow A^2< \left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{200}{201}\right)\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{199}{200}\right)\)
\(\Rightarrow A^2< \frac{1}{201}\left(đpcm\right)\)
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
Ta có :
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\left(đpcm\right)\)
Chúc bạn học tốt !!!
Ta có:
\(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};...;\frac{199}{200}< \frac{200}{201}\)
\(\Rightarrow\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{199}{200}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{200}{201}\)
\(\Rightarrow C< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{200}{201}\)
\(\Rightarrow C^2< \left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{200}{201}\right).\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{199}{200}\right)\)
\(\Rightarrow C^2< \frac{1}{201}\left(dpcm\right)\)
=\(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
=\(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\cdot\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)
=\(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\cdot0\)
\(=0\)
Kết quả = 0 nhé, nhớ ủng hộ mh, mh đang âm diểm
~ HOK TỐT ~
\(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(=\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)
\(=\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\cdot0\)
\(=0\)
