Tìm x, biết x^2-8x+9=0

$x^2-8x+9=0$

$⇔(x^2-8x+16)-7=0$

$⇔(x-4)^2-7=0$

$⇔(x-4+\sqrt{7})(x-4-\sqrt{7})=0$
$⇔\left[ \begin{array}{l}x-4+\sqrt{7}=0\\x-4-\sqrt{7}=0\end{array} \right.⇔\left[ \begin{array}{l}x=4-\sqrt{7}\\x=4+\sqrt{7}\end{array} \right.$

Vậy $\left[ \begin{array}{l}x=4-\sqrt{7}\\x=4+\sqrt{7}\end{array} \right.$.