\(M=1-\frac{2}{x}+\frac{2015}{x^2}\)
Đặt y = 1/x
=> M = 1 - 2y + 2015y2 = 2015.(y2 - 2.y .\(\frac{1}{2015}\) + \(\frac{1}{2015^2}\)) - \(\frac{1}{2015}\) + 1 = 2015.(y - \(\frac{1}{2015}\))2 + \(\frac{2014}{2015}\)\(\ge\) 0 + \(\frac{2014}{2015}\) với mọi y
Vậy Min M = \(\frac{2014}{2015}\) khi y - \(\frac{1}{2015}\) = 0 <=> y = \(\frac{1}{2015}\)<=> 1/x = \(\frac{1}{2015}\) <=> x = 2015