1 reply to this topic

[mayumichan]

Tim GTNN cua A = $x^{2}+\sqrt{x^{4}+\frac{1}{x^{2}}}$

[kanashini]

Đặt $x^2=a(a>0)$

 

$A=a+\sqrt{a^2+\frac{1}{a}}$

 

$\Leftrightarrow \left\{\begin{matrix}A>a & \\(A-a)^2=a^2+\frac{1}{a} &\end{matrix}\right.$

 

$\Leftrightarrow \left\{\begin{matrix}A>a & \\2A.a^2-A^2.a+1=0 &\end{matrix}\right.$

 

$\Delta =(-A^2)^2-4.2A=A^4-8A\geq 0$

 

$A\geq 2$