ta có$:\sqrt{x^2+91}-10=\sqrt{x-2}-1+x^2-9$$\Leftrightarrow \frac{(x-3)(x+3)}{\sqrt{x^2+91}+10}=\frac{x-3}{\sqrt{x-2}+1}+(x-3)(x+3)đổivếrax=3$
ta có$:\sqrt{x^2+91}-10
\geq \sqrt{x-2}-1+x^2-9$$\Leftrightarrow \frac{(x-3)(x+3)}{\sqrt{x^2+91}+10}
\geq \frac{x-3}{\sqrt{x-2}+1}+(x-3)(x+3)
đổivếrax=3$