Question

Tìm MIN \(B=\dfrac{12}{x-1}+\dfrac{x}{3}\)với x > 1

Answer 1

Có B = \(\dfrac{12}{x-1}+\dfrac{x}{3}=\dfrac{12}{x-1}+\dfrac{x-1}{3}+\dfrac{1}{3}\)

Áp dụng BĐT Cô si cho 2 số \(\dfrac{12}{x-1},\dfrac{x-1}{3}>0\)

\(\Rightarrow\dfrac{12}{x-1}+\dfrac{x-1}{3}\ge2.\sqrt{\dfrac{12}{x-1}.\dfrac{x-1}{3}}=4\)

\(\Leftrightarrow\dfrac{12}{x-1}+\dfrac{x-1}{3}+\dfrac{1}{3}\ge\dfrac{13}{3}\)

hay B \(\ge\dfrac{13}{3}\)

Dấu ''='' xảy ra \(\Leftrightarrow\dfrac{12}{x-1}=\dfrac{x-1}{3}\)

\(\Leftrightarrow\left(x-1\right)^2=36\)

\(\Leftrightarrow x-1=6\Leftrightarrow x=7\)

Vậy min B = 13_/3 \(\Leftrightarrow\) x = 7