a) $x^2-6x+5=0$
$⇔x^2-6x+9-4=0$
$⇔(x-3)^2=4$
$⇔\left[ \begin{array}{l}x-3=2\\x-3=-2\end{array} \right.$
$⇔\left[ \begin{array}{l}x=5\\x=1\end{array} \right.$
b) $x^2-3x-7=0$
$⇔x^2-2x.\frac{3}{2}+\frac{9}{4}-\frac{37}{4}=0$
$⇔(x-\frac{3}{2})^2=\frac{37}{4}$
$⇔\left[ \begin{array}{l}x-\frac{3}{2}=\frac{\sqrt[]{37}}{2}\\x-\frac{3}{2}=-\frac{\sqrt[]{37}}{2}\end{array} \right.$
$⇔\left[ \begin{array}{l}x=\frac{3+\sqrt[]{37}}{2}\\x=\frac{3-\sqrt[]{37}}{2}\end{array} \right.$
c) $3x^2-12x+1=0$
$⇔3(x^2-4x+4)-11=0$
$⇔3(x-2)^2=11$
$⇔(x-2)^2=\frac{11}{3}$
$⇔\left[ \begin{array}{l}x-2=\frac{\sqrt[]{33}}{3}\\x-2=-\frac{\sqrt[]{33}}{3}\end{array} \right.$
$⇔\left[ \begin{array}{l}x=\frac{6+\sqrt[]{33}}{3}\\x=\frac{6-\sqrt[]{33}}{3}\end{array} \right.$
d) $x^2+\sqrt[]{2}x-1=0$
$⇔x^2+2x.\frac{1}{\sqrt[]{2}}+\frac{1}{2}-\frac{3}{2}=0$
$⇔(x+\frac{1}{\sqrt[]{2}})^2=\frac{3}{2}$
$⇔\left[ \begin{array}{l}x+\frac{1}{\sqrt[]{2}}=\frac{\sqrt[]{6}}{2}\\x+\frac{1}{\sqrt[]{2}}=-\frac{\sqrt[]{6}}{2}\end{array} \right.$
$⇔\left[ \begin{array}{l}x=\frac{-\sqrt[]{2}+\sqrt[]{6}}{2}\\x=\frac{-\sqrt[]{2}-\sqrt[]{6}}{2}\end{array} \right.$.
