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24-04-2016 - 08:27
cho x, y,z >0. $x\geq z$ tìm min của P = $\frac{xz}{y(y+z)}$ + $\frac{y^{2}}{z(x+y)}$ + $\frac{z}{x+z}$
24-04-2016 - 07:55
cho x, y,z >0. x\geqz tìm min của P = \frac{xz}{y(y+z)} + \frac{y^{2}}{z(x+y)} + \frac{z}{x+z}