`y=x+(4)/(x^2)`
TXD: `D=RR\\{0}`
`y'=1-(8x)/(x^4)`
`=(x^4-8x)/(x^4)`
`y'=0<=>x^4-8x=0`
`<=>x.(x^3-8)=0`
`<=>x=0(ktm),x=2(tm)`
BBT:
\begin{array}{|c|cc|}\hline \text{$x$}&\text{0}&&2&\text{}&+\infty\\\hline \text{$y'$}&\text{}||&-&0&\text{}+&\\\hline \text{$y$}&||\\&\text{}||&\text{}\searrow&&\nearrow\\&||&&3\\\hline\end{array}
Vậy `min_{(0;+\infty)}y=y(2)=3`
Chọn `bbD`
