Tính 

- Hướng dẫn giải

Đáp án cần chọn là: B

$\begin{array}{l}M=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}\mathrm{.....}\frac{{99}^2}{99.100}.\frac{{100}^2}{101}\\ =\frac{1^2{.2}^2{.3}^2{\mathrm{....99}}^2{.100}^2}{\left(1.2\right).\left(2.3\right).\left(3.4\right)\mathrm{....}\left(99.100\right).\left(100.101\right)}\\ =\frac{\left(\mathrm{1.2.3.....99.100}\right).\left(\mathrm{1.2.3....99.100}\right)}{\left(\mathrm{1.2.3.....99.100}\right).\left(\mathrm{2.3....99.100}\right).101}\\ =\frac{1}{101}\end{array}$