$\displaystyle \begin{array}{{>{\displaystyle}l}} B=\frac{x}{x^{2} +20x+100} \ \\ B=\frac{1}{x+20+\frac{100}{x}} \ \\ Áp\ dụng\ bdt\ cosi\ ta\ có\ :\ \\ x+\frac{100}{x} \ \geqslant 2\sqrt{x.\frac{100}{x}} =20\ \\ \rightarrow x+20+\frac{100}{x} \ \geqslant 20+20=40\ \\ Do\ đó\ :\ \frac{1}{x+20+\frac{100}{x}} \leqslant \frac{1}{40}\\ Dấu\ =\ xảy\ ra\ khi\ x=\frac{100}{x} \ \\ \rightarrow x=10\\ Vậy\ MaxB=\frac{1}{40} \ tại\ x=10\ \\ \end{array}$
