Cho hàm số  $f\left(x\right)=\sqrt{x^2-2x+8}$. Giải phương trình  $$$f^{\text{'}}\left(x\right)=-\frac{2}{3}$.

Ta có  $f^{\text{'}}\left(x\right)=\frac{1}{2\sqrt{x^2-2x+8}}.{\left(x^2-2x+8\right)}^{\text{'}}$

 $=\frac{2x-2}{2\sqrt{x^2-2x+8}}=\frac{x-1}{\sqrt{x^2-2x+8}}$.

$f^{\text{'}}\left(x\right)=-\frac{2}{3}\iff \frac{x-1}{\sqrt{x^2-2x+8}}=-\frac{2}{3}$

 $\iff 2\sqrt{x^2-2x+8}=-3\left(x-1\right)\left(x<1\right)$

 $\iff 4\left(x^2-2x+8\right)=9{\left(x-1\right)}^2$

 $\iff 4\left(x^2-2x+8\right)=9\left(x^2-2x+1\right)$

 $\iff 5x^2-10x-23=0$

 $\iff x=\frac{5+2\sqrt{35}}{5}$ (loại);  $x=\frac{5-2\sqrt{35}}{5}$ (nhận).

Vậy   $x=\frac{5-2\sqrt{35}}{5}$.