Question

[Nguyễn Thị Ngọc Hân]

Cho \(y=f\left(x\right)=x^2+\frac{1}{x^2}+2\left(x+\frac{1}{x}\right)+8\) Tìm Min y

Answer 1

[Tạ Ngọc Bảo Anh]

x=yx44444444444444444444444444444

Answer 2

[Kiệt Nguyễn]

Đặt \(x+\frac{1}{x}=t\)thì \(x^2+\frac{1}{x^2}=t^2-2\)

Lúc đó: \(y=f\left(x\right)=t^2-2+2t+8=\left(t^2+2t+1\right)+5=\left(t+1\right)^2+5\ge5\)

Đẳng thức xảy ra khi \(t=x+\frac{1}{x}=-1\Leftrightarrow x^2+x+1=0\Leftrightarrow\left(x+\frac{1}{2}\right)^2=-\frac{3}{4}\)\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=\frac{\sqrt{3}}{2}i\\x+\frac{1}{2}=-\frac{\sqrt{3}}{2}i\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{3}i-1}{2}\\x=\frac{-\sqrt{3}i-1}{2}\end{cases}}\)