Question

\(D=\frac{-1^2}{1.2}.\frac{-2^2}{3.2}...\frac{-101^2}{101.102}>\frac{-1}{100}\)

Answer 1

\(D=\frac{\left(-1\right).\left(-1\right)}{1.2}.\frac{\left(-2\right).\left(-2\right)}{2.3}...\frac{\left(-101\right).\left(-101\right)}{101.102}\)

\(=\frac{\left(-1\right)\left(-1\right)\left(-2\right)\left(-2\right)...\left(-101\right)\left(-101\right)}{1.2.2.3...101.102}\)

\(=\frac{\left[\left(-1\right)\left(-2\right)...\left(-101\right)\right].\left[\left(-1\right).\left(-2\right)...\left(-101\right)\right]}{\left(1.2...101\right).\left(2.3...102\right)}\)

\(=\left(-1\right).\frac{-1}{102}\)

\(=\frac{1}{102}\)

Vì \(\frac{1}{102}>\frac{-1}{100}\)

Vậy\(D>\frac{-1}{100}\)