\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\)
\(=\frac{1}{100}\)
\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)
\(=\frac{1}{100}\)
+)30 c4^`n 71k 0k
\(\frac{1}{2}.\frac{2}{3}...\frac{98}{99}.\frac{99}{100}\)
\(=\frac{1.2...98.99}{2.3...99.100}\)
\(=\frac{1}{100}\)
\(\frac{1}{2}x\frac{2}{3}x...x\frac{98}{99}x\frac{99}{100}=\frac{1}{100}\)
Ta có :
\(\frac{1}{2}.\frac{2}{3}....\frac{98}{99}.\frac{99}{100}\)
\(=\frac{1.2....98.99}{2.3....99.100}\)
\(=\frac{1}{100}\)
\(\frac{1.2.3....98.99}{2.3.4....99.100}=\frac{1}{100}\)
\(\frac{1}{2}.\frac{2}{3}....\frac{98}{99}.\frac{99}{100}\)
\(=\frac{1.2....98.99}{2.3...99.100}=\frac{1}{100}\)
Vậy kết quả cần tìm là : \(\frac{1}{100}\)
\(\frac{1}{2}.\frac{2}{3}.....\frac{98}{99}.\frac{99}{100}\)
\(=\frac{1.2.3...98.99}{2.3....99.100}\)
\(=\frac{1}{100}\)