Question

Tìm GTNN của biểu thức \(G=\frac{9x}{2-x}+\frac{2}{x}\)

Answer 1

Theo bất đẳng thức Cauchy :

\(G=\frac{9x}{2-x}+\frac{2-x}{x}+1\ge2\sqrt{\frac{9x\left(2-x\right)}{\left(2-x\right)x}}+1=7\)

Đẳng thức xảy ra khi ...

tự tìm dấu = :))

Answer 2

Trả lời:

\(G=\frac{9}{2-x}+\frac{2}{x}\)\(\left(ĐK:0< x< 2\right)\)

\(G=\frac{9}{2-x}+\frac{2-x+x}{x}\)

\(G=\frac{9}{2-x}+\frac{2-x}{x}+1\)

Áp dụng BĐT Cauchy ta có:

\(\frac{9x}{2-x}+\frac{2-x}{x}\ge2.\sqrt{\frac{9x}{2-x}\times\frac{2-x}{x}}=2.3=6\)

\(\Leftrightarrow\frac{9}{2-x}+\frac{2-x}{x}+1\ge6+1=7\)

Hay \(G\ge7\)

Dấu "=" xảy ra \(\Leftrightarrow\frac{9x}{2-x}=\frac{2-x}{x}\)

\(\Leftrightarrow\left(2-x\right)^2=9x^2=\left(\pm3x\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}2-x=3x\\2-x=-3x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2=4x\\2=-2x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\left(TM\right)\\x=-1\left(L\right)\end{cases}}\)

Vậy \(G_{min}=7\Leftrightarrow x=\frac{1}{2}\)