BT1:Cho x>=3 ,tìm min f(x)= x+ 1/x
BT2: Cho x>=2 ,tìm min f(x)= x + 1/x^2
Tks.......................
đây là dạng bài tìm điểm rơi
câu 1
[laTEX]A = x +\frac{1}{x} \\ \\ A = \frac{x}{9} + \frac{1}{x} + \frac{8x}{9} \geq 2.\sqrt{\frac{x}{9}.\frac{1}{x}} + \frac{8.3}{9} = \frac{10}{3} \\ \\ Min A =\frac{10}{3} \\ \\ x = 3 [/laTEX]
câu 2
[laTEX]B = x + \frac{1}{x^2 }\\ \\ B = \frac{x}{8} + \frac{x}{8} + \frac{1}{x^2} + \frac{3x}{8} \geq 3.\sqrt[3]{\frac{x}{8}.\frac{x}{8}.\frac{1}{x^2} } + \frac{3.2}{8} = \frac{9}{4} \\ \\ Min B = \frac{9}{4} \\ \\ x = 2 [/laTEX]
Last edited by a moderator: