Recent Content by tienminh1998

Gọi AB = a, BC= b. Do $I= \dfrac{P}{4\pi R^{2}}$ $\Rightarrow$ I max khi R min Ta có $AM= \sqrt{\dfrac{AB^{2}}{4}+BC^{2}}$ Mặt khác $\dfrac{a^{2}}{4}+b^{2}\geq ab$ Suy ra $AM_{min}=\sqrt{18}$ $A'M_{min}= \sqrt{18+3^{2}}=\sqrt{27}$ Ta có $I_{1}+I_{2}+I_{3}+I_{4}=10$ $\Leftrightarrow$...