Cho C = \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{199}{200}\) Cm C2 < \(\frac{1}{201}\)
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CM \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{199}-\frac{1}{200}=\frac{1}{101}+\frac{1}{102}+....+\frac{1}{200}\)
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
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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 !!!
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Cho A=\(\frac{2}{1}.\frac{4}{3}.\frac{6}{5}...\frac{200}{199}CMR:14< A< 20\)
cho \(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{199}{200}\)
chứng minh :\(C^2
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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Xem thêm câu trả lời
C1. Tìm frac {a}b, biết rằng :a, Afrac{1}{2}+frac{1}{3}+frac{1}{4}+...+frac{1}{200}b, Bfrac{1}{199}+frac{2}{198}+frac{3}{197}+...+frac{198}{2}+frac{199}{1}c, Cfrac{1}{1cdot2cdot3cdot4}+frac{1}{2cdot3cdot4cdot5}+...+frac{1}{27cdot28cdot29cdot30}C2. Tìm x :1. 1+frac{1}{3}+frac{1}{6}+frac{1}{10}+...+frac{1}{frac{xleft(x+1right)}{2}}1frac{1991}{1993}2. frac{1}{5times8}+frac{1}{8times11}+...+frac{1}{11times14}+...+frac{1}{xtimes(x+3)}frac{101}{1540}
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C1. Tìm \(\frac {a}b\), biết rằng :
a, \(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}\)
b, \(B=\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}\)
c, \(C=\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+...+\frac{1}{27\cdot28\cdot29\cdot30}\)
C2. Tìm x :
1. \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{\frac{x\left(x+1\right)}{2}}=1\frac{1991}{1993}\)
2. \(\frac{1}{5\times8}+\frac{1}{8\times11}+...+\frac{1}{11\times14}+...+\frac{1}{x\times(x+3)}=\frac{101}{1540}\)
Tìm số tự nhiên x,y biết :x-3=y*(x+2)
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CMR: \(14< \frac{2}{1}.\frac{4}{3}.\frac{6}{5}....\frac{200}{199}< 20\)
Cho A=\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{199}{200}\).CMR \(A^2\)<\(\frac{1}{201}\)
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 \(A^2< \frac{1}{201}\)
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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 \(S=\frac{2}{1}.\frac{4}{3}.\frac{6}{5}.\frac{8}{7}.......\frac{200}{199}\)
CMR: 201<S2<400
Giải đúng và chi tiết mk **** cho