Question

[Azenda]

Với x>3. Hãy tìm GTNN của M=(x^2+2x-9)/(x-3)

Answer 1

[River flows in you]

\(M=\frac{x^2+2x-9}{x-3}\)

\(=\frac{x^2-6x+9+8x-24+6}{x-3}\)

\(=\frac{\left(x-3\right)^2+8\left(x-3\right)+6}{x-3}\)

\(=x-3+8+\frac{6}{x-3}\)

Do \(x>3\Rightarrow x-3>0\)

Áp dụng BĐT Cauchy , ta có : 

\(x-3+\frac{6}{x-3}\ge2\sqrt{\left(x-3\right).\frac{6}{x-3}}=2\sqrt{6}\)

\(\Rightarrow M=x-3+\frac{6}{x-3}+8\ge2\sqrt{6}+8\)

\(\Rightarrow M\ge\sqrt{24}+8\)

Dấu " = " xảy ra \(\Leftrightarrow x-3=\frac{6}{x-3}\Leftrightarrow\left(x-3\right)^2=6\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=\sqrt{6}\\x-3=-\sqrt{6}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3+\sqrt{6}\left(TM\right)\\x=3-\sqrt{6}\left(L\right)\end{cases}}}\)

Vậy Min M là : \(\sqrt{24}+8\Leftrightarrow x=3+\sqrt{6}\)