Question

C= x + \(\dfrac{1}{x}\) với x > hoặc bằng 4

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PHƯƠNG PHÁP ĐIỂM DƠI

Answer 1

\(C=\dfrac{1}{x}+\dfrac{x}{16}+\dfrac{15}{16}x\ge2\sqrt{\dfrac{1}{x}.\dfrac{x}{16}}+\dfrac{15}{16}.4=\dfrac{1}{2}+\dfrac{15}{4}=\dfrac{17}{4}\)

dấu = xảy ra khi x=4

Answer 2

\(x+\dfrac{1}{x}=\dfrac{1}{16}x+\dfrac{1}{x}+\dfrac{15}{16}x\ge2\sqrt{\dfrac{x}{16x}}+\dfrac{15}{16}.4=\dfrac{1}{2}+\dfrac{15}{4}=\dfrac{17}{4}\)

\(minC=\dfrac{17}{4}\Leftrightarrow x=4\)

Answer 3

\(C=x+\dfrac{1}{x}=\dfrac{x}{16}+\dfrac{1}{x}+\dfrac{15x}{16}\ge2\sqrt{\dfrac{1}{16}}+\dfrac{15.4}{16}=\dfrac{17}{4}\)

dấu"=" xảy ra<=>x=4