Câu hỏi của Kiều Quốc Nam

Question

$D=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot......\cdot\frac{99}{100}$ chứng minh rằng $D< \frac{1}{10}$

soyeon_Tiểu bàng giải · Accepted Answer

$D=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}$$D< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}$$D^2< \frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}...\frac{99}{100}.\frac{100}{101}$$D^2< \frac{1}{101}< \frac{1}{100}=\left(\frac{1}{10}\right)^2$=> $D< \frac{1}{10}\left(đpcm\right)$Câu trả lời: $D=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}$$D< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}$$D^2< \frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}...\frac{99}{100}.\frac{100}{101}$$D^2< \frac{1}{101}< \frac{1}{100}=\left(\frac{1}{10}\right)^2$=> $D< \frac{1}{10}\left(đpcm\right)$

oOo _ Virgo _ oOo · Answer

$D=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}$$D< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}$$D^2< \frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}...\frac{99}{100}.\frac{100}{101}$$D^2< \frac{1}{101}< \frac{1}{100}=\left(\frac{1}{10}\right)^2$$= >D< \frac{1}{10}$$\text{k tui}$Câu trả lời: $D=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}$$D< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}$$D^2< \frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}...\frac{99}{100}.\frac{100}{101}$$D^2< \frac{1}{101}< \frac{1}{100}=\left(\frac{1}{10}\right)^2$$= >D< \frac{1}{10}$$\text{k tui}$

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