So sánh M và N, biết
\(M=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\)và \(N=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\)
Cho M=\(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot....\cdot\frac{99}{100}\)
N=\(\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot......\cdot\frac{100}{101}\)
a, Tính M\(\times\)N
b, CM M<\(\frac{1}{10}\)
Bài 19, Cho \(A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\)
\(B=\frac{1}{10}\)
So sánh A và B
Bài 20, Cho \(A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{999}{1000}\)
\(B=\frac{1}{100}\)
So sánh A và B
Bài 21, Cho \(A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{2499}{2500}\)
Chứng minh A<\(\frac{1}{49}\)
Bài 20, Cho \(A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\)
\(B=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\)
\(C=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{98}{99}\)
1/ So sánh A, B, C
2/Chứng minh \(A\cdot C< A^2< \frac{1}{10}\)
3/Chứng minh \(\frac{1}{15}< A< \frac{1}{10}\)
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{98}{99}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{98}{99}.\frac{99}{100}\)
\(\Rightarrow A^2>\frac{1}{100}=\frac{1}{10^2}\)
Vậy \(A>\frac{1}{10}\)
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9999}{10000}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{9998}{9999}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{9998}{9999}.\frac{9999}{10000}\)
\(\Rightarrow A^2>\frac{1}{10000}=\frac{1}{100^2}\)
\(VayA>\frac{1}{100}=B\)
Bài 22, Cho \(A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\)
\(B=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\)
1/ So sánh A và B, A2 và A.B
2/ Chứng minh A<\(\frac{1}{10}\)
Bài 21, Cho \(A=\frac{1\cdot3\cdot5\cdot...\cdot4095}{2\cdot4\cdot6\cdot...\cdot4096}\)
\(B=\frac{2\cdot4\cdot6\cdot...\cdot4096}{1\cdot3\cdot5\cdot...\cdot4097}\)
1/ So sánh A2 và A.B
2/ Chứng minh A<\(\frac{1}{64}\)
Bài 21, Cho \(A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{2499}{2500}\)Chứng minh A<\(\frac{1}{49}\)
Bài 22, Cho \(A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\)
\(B=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\)
\(C=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{98}{99}\)
1/ So sánh A, B, C
2/Chứng minh \(A\cdot C< A^2< \frac{1}{10}\)
3/Chứng minh \(\frac{1}{15}< A< \frac{1}{10}\)
Chứng minh rằng:
\(A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\frac{7}{8}\cdot...\cdot\frac{99}{100}
Đặt \(B=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.\frac{8}{9}....\frac{100}{101}\)
Nhận xét: Nếu \(\frac{a}{b}
Chứng minh \(A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\frac{7}{8}\cdot...\cdot\frac{99}{100}<\frac{1}{\sqrt{151}}\)
So Sánh A=\(\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\cdot\cdot\cdot\frac{4998}{4999}\)và 0,02
Chứng minh rằng:
\(A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\frac{7}{8}\cdot...\cdot\frac{99}{100}<\frac{1}{\sqrt{151}}\)
so sánh A và B biết:
B=\(\frac{1}{100}\)
A=\(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{9999}{10000}\)
Đặt C = \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}\)\(\left(C>0\right)\)
Và D = \(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{10000}{10001}\)\(\left(D>0\right)\)
Ta có :
C .D = \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{10000}{10001}\)\(=\frac{1}{10001}\)\(\left(1\right)\)
Mặt khác :
\(\frac{1}{2}< \frac{2}{3}\)
\(\frac{3}{4}< \frac{4}{5}\)
\(.....\)
\(\frac{9999}{10000}< \frac{10000}{10001}\)
Nhân tất cả vế theo vế - - - > C < D - - - > C2 < C . D \(\left(2\right)\)
\(\left(1\right),\left(2\right)\)- - - >C2 < \(\frac{1}{10001}\)- - - > C < căn \(\frac{1}{10001}\)< căn \(\frac{1}{10000}\)= \(\frac{1}{100}\)( đpcm )
\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot...\cdot\frac{99}{100}\)
Chúng ta vừa làm vừa triệt tiêu
Cuối cùng còn 1/100
\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{99}{100}=\frac{1.2.3.4...99}{2.3.4.5...100}=\frac{1}{100}\)