类型一 用配方法解方程
【典例 1】$5(x-3)^2=125$.
变式.$x(x+2)=1$.

【典例 1】解：$(x-3)^2=25$,
$x-3=\pm5$,
$\therefore x_1=8,x_2=-2$.
变式.解:$x^2+2x=1$,
$(x+1)^2=2$,
$x=-1\pm\sqrt{2}$,
$\therefore x_1=-1+\sqrt{2},x_2=-1-\sqrt{2}$.

【典例2】$x^2 + x - 1 = 0$.
变式.$x^2 + 3x - 1 = 0$.
$ax^2 + bx + c = 0(a≠0)$
两根为 $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{$$}$ ($b^2 - 4ac ≥ 0$)

【典例 2】解:$\Delta = b^2 - 4ac = 5$,
$x_1=\frac{-1+\sqrt{5}}{2}$,
$x_2=\frac{-1-\sqrt{5}}{2}$.
变式.解:$\Delta=9+4=13$,
$x_1=\frac{-3+\sqrt{13}}{2}$,
$x_2=\frac{-3-\sqrt{13}}{2}$.

【典例3】$x(x-2)=3x-6.$
变式.$x^2 -6x=-9.$
$ax^2 +bx +c=0$
解:$a(x-x$$(x-x_2)=0$
$\therefore x=x_1,x=x_2$

【典例 3】解:$(x-2)(x-3)=0$,
$x_1=2,x_2=3$.
变式.解:$(x-3)^2=0$,
$\therefore x_1=x_2=3$.

【典例4】(1)$2x^2 - 4x = 0$；(2)$(2x - 1)(x + 3) = 4$.
变式.(1)$3x(x - 1) = 2(x - 1)$；(2)$(2x - 1)^2 = (3 - x)^2$.

【典例 4】解:(1)$2x(x-2)=0$,
$\therefore x_1=0,x_2=2$.
(2)$2x^2+5x-7=0$,
$\Delta=5^2-4×2×(-7)=81$,
$x=\frac{-5\pm9}{4}$,
$\therefore x_1=-\frac{7}{2},x_2=1$.
变式.解:(1)$(x-1)(3x-2)=0$,
$\therefore x_1=1,x_2=\frac{2}{3}$.
(2)$(2x-1+3-x)(2x-1+x-3)=0$,
$(x+2)(3x-4)=0$,
$\therefore x_1=\frac{4}{3},x_2=-2$.