cho \(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{199}{200}\)
chứng minh :\(C^2
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CHO C=\(\frac{1}{2}\).\(\frac{3}{4}\).\(\frac{5}{6}\)....\(\frac{199}{200}\)
CHỨNG MINH RẰNG \(C^2\)<\(\frac{1}{201}\)
Ta có:
\(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};...;\frac{199}{200}< \frac{200}{201}\)
\(\Rightarrow\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{199}{200}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{200}{201}\)
\(\Rightarrow C< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{200}{201}\)
\(\Rightarrow C^2< \left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{200}{201}\right).\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{199}{200}\right)\)
\(\Rightarrow C^2< \frac{1}{201}\left(dpcm\right)\)
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Cho \(A=\frac{1}{2}\times\frac{3}{4}\times\frac{5}{6}\times...\times\frac{199}{200}\)và chứng minh \(A^2< \frac{1}{201}\)
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)
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Ta có:
\(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};...;\frac{199}{200}< \frac{200}{201}\)
\(\Rightarrow\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{199}{200}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{200}{201}\)
\(\Rightarrow A< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{200}{201}\)
\(\Rightarrow A^2< \left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{200}{201}\right)\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{199}{200}\right)\)
\(\Rightarrow A^2< \frac{1}{201}\left(đpcm\right)\)
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\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{199}{200}\)
\(\Rightarrow A< \frac{2}{3}.\frac{4}{5}\frac{6}{7}...\frac{200}{201}\)
\(\Rightarrow A.A< A.\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{200}{201}\right)\)
\(\Rightarrow A^2< \frac{1}{201}\)(làm phần trc như Sakuraba Laura nhá)
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Cho \(S=\frac{1}{2}+\frac{3}{4}+\frac{5}{6}+...+\frac{199}{200}\)Chứng minh rằng : \(S^2<\frac{1}{201}\)
chứng minh rằngBfrac{1}{2}*frac{3}{4}*frac{5}{6}*...*frac{199}{200}.CHỨNG MINH B2frac{1}{201}C 1+frac{1}{2}+frac{1}{3}+frac{1}{4}+...+frac{1}{2^{100}-1}CHỨNG MINH C100CÍU MÌNH CHIỀU NỌP RỒI ẠTHANK YOU
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chứng minh rằng
B=\(\frac{1}{2}\)*\(\frac{3}{4}\)*\(\frac{5}{6}\)*...*\(\frac{199}{200}\).CHỨNG MINH B2<\(\frac{1}{201}\)
C= 1+\(\frac{1}{2}\)+\(\frac{1}{3}\)+\(\frac{1}{4}\)+...+\(\frac{1}{2^{100}-1}\)CHỨNG MINH C<100
CÍU MÌNH CHIỀU NỌP RỒI Ạ
THANK YOU
Chứng minh: \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{199}+\frac{1}{200}\)
Ta có : \(1-\frac{1}{2}+\frac{1}{3}-...-\frac{1}{200}=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)\(\left(đpcm\right)\)
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Cho C = \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{199}{200}\) Cm C2 < \(\frac{1}{201}\)
( 2 cách nha )
Chứng minh rằng
a)\(\frac{5}{8}<\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}<\frac{3}{4}\)
b)\(\frac{1}{4}+\frac{1}{6}+\frac{1}{16}+...+\frac{1}{10000}<\frac{3}{4}\)
c)(\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{199}{200}\))2 <\(\frac{1}{201}\)
d)\(50<1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}...+\frac{1}{2^{100}-1}<100\)
Chứng minh :
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
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https://olm.vn/hoi-dap/detail/5631756599.html
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CMR
\(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{199}{200}\)
Có C^2 < 1/201
C = 1/200
=> C^2 = 1/400 < 1/201
=> C^2 < 1/201 (đpcm)
K nhé!
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Ta rút gọn C = 1/200
=> C^2 = 1/400
Mà 1/400 < 1/201
=> C^2 < 1/201 (đpcm)
Ai k mk mk k lại !!
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Ta rút gọn C = 1/200
=> C^2 = 1/400
Mà 1/400 < 1/201
=> C^2 < 1/201 (đpcm)
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