tìm giá trị nhỏ nhất của (x^2-2x+206)/x^2

I

iceghost

$\dfrac{x^2-2x+206}{x^2} \\
=\dfrac{\dfrac{x^2}{206}-2x+206+\dfrac{205x^2}{206}}{x^2} \\
=\dfrac{(\dfrac{x}{\sqrt{206}})^2-2.\dfrac{x}{\sqrt{206}}.\sqrt{206}+(\sqrt{206})^2}{x^2}+\dfrac{205}{206} \\
=\dfrac{(\dfrac{x}{\sqrt{206}}-\sqrt{206})^2}{x^2}+\dfrac{205}{206} \ge \dfrac{205}{206} \\
\implies Min = \dfrac{205}{206} \iff \dfrac{x}{\sqrt{206}}-\sqrt{206} = 0 \iff x=206$
 
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