Cho \(A=\frac{2}{1}\times\frac{3}{2}\times\frac{6}{5}\times...\times\frac{200}{199}\)
CMR: A < 20
CT
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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 A =\(\frac{1}{2}\times\frac{3}{4}\times\frac{5}{6}\times\frac{7}{8}\times....\times\frac{199}{200}\)
Chứng minh A2 < \(\frac{1}{400}\)
GIÚP MÌNH VỚI CÁC BẠN
1. Tính nhanh:a.frac{17}{13}timesfrac{7}{15}-frac{5}{12}timesfrac{21}{39}+frac{49}{91}timesfrac{8}{15}b.left(frac{12}{199}+frac{23}{200}-frac{34}{201}right)timesleft(frac{1}{2}-frac{1}{3}-frac{1}{6}right)2. So sánh:a. 3200và2300b. 7150và3775c.frac{201201}{202202}vàfrac{201201201}{202202202}
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1. Tính nhanh:
a.\(\frac{17}{13}\times\frac{7}{15}-\frac{5}{12}\times\frac{21}{39}+\frac{49}{91}\times\frac{8}{15}\)
b.\(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\times\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
2. So sánh:
a. 3200và2300
b. 7150và3775
c.\(\frac{201201}{202202}\)và\(\frac{201201201}{202202202}\)
2. a) \(3^{200}=\left(3^2\right)^{100}=9^{100}\)
\(2^{300}=\left(2^3\right)^{100}=8^{100}\)
Vì \(9^{100}>8^{100}\Rightarrow3^{200}>2^{300}\)
b) \(71^{50}=\left(71^2\right)^{25}=5041^{25}\)
\(37^{75}=\left(3^3\right)^{25}=27^{25}\)
Vì \(5041^{25}>27^{25}\Rightarrow71^{50}>37^{75}\)
c) \(\frac{201201}{202202}=\frac{201201:1001}{202202:1001}=\frac{201}{202}\)
\(\frac{201201201}{202202202}=\frac{201201201:1001001}{202202202:1001001}=\frac{201}{202}\)
Vì \(\frac{201}{202}=\frac{201}{202}\Rightarrow\frac{201201}{202202}=\frac{201201201}{202202202}\)
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A = \(\frac{1}{2}\times\frac{3}{4}\times\frac{5}{6}\times...\times\frac{79}{80}.\)CMR: A bé hơn \(\frac{1}{9}\)
Xem chi tiết
ta có 1/2 * 3/ 4 * 5/6 *... * 79/80 = 0.0889
so sánh a với 1/9
0.0889 < 0.(1)
=> A < 1/9
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CMR: 1 x 3 x 5 x 7 x...x 19 = \(\frac{11}{2}\times\frac{12}{2}\times\frac{13}{2}\times...\times\frac{20}{2}\)
Xem chi tiết
Ta có: \(1.3.5.7....19=\frac{1}{1}.\frac{3}{1}.\frac{5}{1}.\frac{7}{1}....\frac{19}{1}\)
Mà \(1.3.5.7....19=\frac{11.12.13....20}{2.2.2....2}\)
\(\Rightarrow\frac{1}{1}.\frac{3}{1}.\frac{5}{1}.\frac{7}{1}....\frac{19}{1}=\frac{11.12.13....20}{2.2.2...2}\)
\(\Rightarrow1.3.5.7...19=\frac{11}{2}.\frac{12}{2}.\frac{13}{2}.....\frac{20}{2}\)(đpcm)
P/s: Mấy bọn ko biết giải thì câm mồm vào đừng chọn sai nha!!! (Mình không nói bạn Đức Minh Nguyễn nha)
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1/tìm STN nhỏ nhất chia cho 5 dư 1,chia7 dư 5 2/CMR: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} 3/CMR:frac{51}{2}timesfrac{52}{2}times...timesfrac{100}{2}1times3times5times...times97times99 4/cho Afrac{1}{2}timesfrac{3}{4}timesfrac{5}{6}times...timesfrac{9999}{...
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1/tìm STN nhỏ nhất chia cho 5 dư 1,chia7 dư 5 2/CMR:\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{199}-\frac{1}{200}=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\) 3/CMR:\(\frac{51}{2}\times\frac{52}{2}\times...\times\frac{100}{2}=1\times3\times5\times...\times97\times99\) 4/cho A=\(\frac{1}{2}\times\frac{3}{4}\times\frac{5}{6}\times...\times\frac{9999}{10000}\) so sánh A với 0,01 5/CMR:\(\left(1+2+3+...+n\right)-7\) chia hết cho 10
Cho biểu thức A= \(\frac{2}{1}\times\frac{4}{3}\times\frac{6}{5}\times\frac{8}{7}\times\frac{10}{9}\times...\times\frac{100}{99}\)Chứng minh rằng 12<A<13
Chứng minh rằng:\(\frac{-1}{2}\times\frac{-3}{4}\times\frac{-5}{6}\times...\times\frac{-399}{400}< \frac{1}{20}\)
em hỏi thầy cô đây là toán chuyên
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tính bằng cách thuận tiện
a. \(\frac{1}{2}\times\frac{2}{3}\div\frac{4}{3}\times\frac{4}{5}\div\frac{6}{5}\times\frac{6}{7}\div\frac{7}{8}\times\frac{8}{9}\div\frac{10}{9}\)
b.\(\frac{27}{49}\times\frac{49}{50}\times\frac{15}{51}\times(\frac{5}{10}-\frac{1}{2})\)
(7x6=5+2+6x7) =