\(y=\sqrt[3]{x^4+16x^2+64}-3\sqrt[3]{x^2+8}+1\)
\(=\sqrt[3]{(x^2+8)^2}-3\sqrt[3]{x^2+8}+1\)
Đặt
\(\sqrt[3]{x^2+8}=t\Rightarrow t\geq \sqrt[3]{8}=2\)
Khi đó:
\(y=t^2-3t+1=t(t-2)-(t-2)-1\)
\(=(t-1)(t-2)-1\)
Vì
\(t\geq 2\Rightarrow (t-1)(t-2)\geq 0\Rightarrow y\geq 0-1=-1\)
`text{Vậy Min_y=-1 khi x=0}`