Question

Cho A =  \(\frac{1}{2}\)x   \(\frac{3}{4}\)x   \(\frac{5}{6}\)x ... x   \(\frac{99}{100}\)và  B =   \(\frac{2}{3}\)x   \(\frac{4}{5}\)x   \(\frac{6}{7}\)x ... x   \(\frac{100}{101}\)

a) CMR: A < B

b) Tính A x B

Answer 1

a) Vì \(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};...;\frac{99}{100}< \frac{100}{101}\)nên:

\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)

hay A < B (đpcm)

b) \(AB=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}...\frac{99}{100}.\frac{100}{101}\)

\(\Leftrightarrow AB=\frac{1.2.3.4.5.6...99.100}{2.3.4.5.6.7....100.101}\)

\(\Leftrightarrow AB=\frac{1}{101}\)

Vậy \(AB=\frac{1}{101}\)

Answer 2

a, So sánh từng nhân tử của hai vế ta thấy:

\(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};...;\frac{99}{100}< \frac{100}{101}\)

Suy ra \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)

Suy ra A<B

b, \(A.B=\frac{1.2.3.4.5.6...99.100}{2.3.4.5.6.7...100.101}=\frac{1}{101}\)