\(\text{Cho }P=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot\cdot\cdot\frac{399}{400}\text{ Chứng minh }P< \frac{1}{20}\)
P=\(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\cdot\cdot\cdot\frac{399}{400}\)
Chứng tỏ rằng \(P< \frac{1}{20}\)
\(P=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{399}{400}\)
\(\Rightarrow P< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{400}{401}\)
\(\Rightarrow P^2< \frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}...\frac{399}{400}.\frac{400}{401}\)
\(\Rightarrow P^2< \frac{1}{401}< \frac{1}{400}=\frac{1}{20^2}\)
\(\Rightarrow P< \frac{1}{20}\)
P=1/2.3/4.5/6.....399/400
=>P<2/3.4/5......400/401
=>P2<1/2.2/3.3/4......398/399.399/400.400/401
=1/401<1/400=(1/20)2
=>P<1/20
thanks vì trả lời câu hỏi tớ nhá
NGUYỄN TUẤN CHÂU KIỆT
cho A=\(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\cdot\cdot\frac{99}{100}\)
CHỨNG MINH \(\frac{1}{15}< a< \frac{1}{10}\)
1. Tính tổng
\(\frac{1}{2}\cdot\frac{1}{3}\cdot+\frac{1}{3}\cdot\frac{1}{4}+\frac{1}{4}\cdot\frac{1}{5}+\frac{1}{5}\cdot\frac{1}{6}+\frac{1}{6}\cdot\frac{1}{7}+\frac{1}{7}\cdot\frac{1}{8}+\frac{1}{8}\cdot\frac{1}{9}\)
\(\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{4}+\frac{1}{4}\cdot\frac{1}{5}+\frac{1}{5}\cdot\frac{1}{6}+\frac{1}{6}\cdot\frac{1}{7}+\frac{1}{7}\cdot\frac{1}{8}+\frac{1}{8}\cdot\frac{1}{9}\)
\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{2}-\frac{1}{9}=\frac{7}{18}\)
\(\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{4}+...+\frac{1}{8}\cdot\frac{1}{9}\)
\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{2}-\frac{1}{9}\)
* LÀM NỐT *
#Louis
1/2.1/3+1/3.1/4+1/4.1/5+...+1/8.1/9
=1/2.3=1/3.4+1/4.5+...+1/8.9\
=1/2-1/3+1/3-1/4=1/4-1/5+...+1/8.1/9
=1/2-1/9
=9/18-2/18
=7/18
HỌC TỐT NHA BẠN
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}
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 \(A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\frac{7}{8}\cdot...\cdot\frac{99}{100}<\frac{1}{\sqrt{151}}\)
Chứng minh rằng:
\(A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\frac{7}{8}\cdot...\cdot\frac{99}{100}<\frac{1}{\sqrt{151}}\)
CHO A = \(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\cdot\cdot\frac{99}{100}\)
CHỨNG MINH \(\frac{1}{15}< A< \frac{1}{10}\)
cho:
\(A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot.........\cdot\frac{2017}{2018}\)
chứng minh A<\(\frac{1}{2018}\)