Có : \(P=4x+\dfrac{16}{x^2}=2x+2x+\dfrac{16}{x^2}\)
- AD AMGM : \(2x+2x+\dfrac{16}{x^2}\ge3\sqrt[3]{\dfrac{2x.2x.16}{x^2}}=12\)
- Dấu " = " xảy ra \(\Leftrightarrow2x=\dfrac{16}{x^2}\)
\(\Leftrightarrow2x^3=16\)
\(\Leftrightarrow x=2\) ( TM )
Vậy ....
( Chắc đề như vầy :vvv )
Hoặc là như này :vvv
\(P=\dfrac{4x+16}{x^2}=\dfrac{4}{x}+\left(\dfrac{4}{x}\right)^2=\left(\dfrac{4}{x}\right)^2+\dfrac{2.4}{x}.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\dfrac{4}{x}+\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)
- Thấy : \(\left(\dfrac{4}{x}+\dfrac{1}{2}\right)^2\ge0\)
\(\Rightarrow P\ge-\dfrac{1}{4}\)
- Dấu " = " xảy ra \(\Leftrightarrow x=-8\left(L\right)\)
Vậy ....
( Chắc đề này sai r :vv )
