《2025年名校课堂九年级数学上册人教版江西专版》

1. 用指定的方法解一元二次方程：
(1)$64(x + 1)^2 = 81$。（直接开平方法）
(2)$x^2 - 4x + 2 = 0$。（配方法）
(3)$x(x - 3) = 10(x - 3)$。（因式分解法）
(4)$2x^2 - 7x - 2 = 0$。（公式法）

答案： 1.解:
(1)$(x+1)^{2}=\frac {81}{64},\therefore x+1=\pm \frac {9}{8}.\therefore x_{1}=\frac {1}{8},x_{2}=-\frac {17}{8}$.
(2)$x^{2}-4x+4=2,(x-2)^{2}=2,\therefore x-2=\pm \sqrt {2}.\therefore x_{1}=2+\sqrt {2},x_{2}=2-\sqrt {2}$.
(3)$x(x-3)-10(x-3)=0,(x-3)(x-10)=0,\therefore x-3=0$或$x-10=0.\therefore x_{1}=3,x_{2}=10$.
(4)$\because a=2,b=-7,c=-2,\therefore \Delta =(-7)^{2}-4×2×(-2)=65＞0.\therefore x=\frac {7\pm \sqrt {65}}{2×2}=\frac {7\pm \sqrt {65}}{4}.\therefore x_{1}=\frac {7+\sqrt {65}}{4},x_{2}=\frac {7-\sqrt {65}}{4}.$

2. 用合适的方法解一元二次方程：
(1)$(2x + 3)^2 = (3x + 2)^2$。
(2)$4x(x - 6) = 1$。
(3)$1 - 8x + 16x^2 = 2 - 8x$。
(4)$3(x - 2)^2 = x^2 - 4$。
(5)$(x + 1)(x + 3) = 2$。
(6)$(2x + 1)^2 = x^2 + 2$。

答案： 2.解:
(1)开平方,得$2x+3=3x+2$或$2x+3=-3x-2$.解得$x_{1}=1,x_{2}=-1$.
(2)$x^{2}-6x=\frac {1}{4}.x^{2}-6x+9=\frac {1}{4}+9$,即$(x-3)^{2}=\frac {37}{4}.\therefore x-3=\frac {\sqrt {37}}{2}$或$x-3=-\frac {\sqrt {37}}{2}.\therefore x_{1}=3+\frac {\sqrt {37}}{2},x_{2}=3-\frac {\sqrt {37}}{2}$.
(3)$16x^{2}=1,x^{2}=\frac {1}{16},\therefore x=\pm \frac {1}{4}.\therefore x_{1}=\frac {1}{4},x_{2}=-\frac {1}{4}$.
(4)$3(x-2)^{2}-(x+2)(x-2)=0.(x-2)(3x-6-x-2)=0.\therefore x-2=0$或$2x-8=0.\therefore x_{1}=2,x_{2}=4$.
(5)整理,得$x^{2}+4x=-1.x^{2}+4x+4=-1+4$,即$(x+2)^{2}=3.\therefore x+2=\pm \sqrt {3}.\therefore x_{1}=-2+\sqrt {3},x_{2}=-2-\sqrt {3}$.
(6)整理,得$3x^{2}+4x-1=0.\because a=3,b=4,c=-1,\therefore \Delta =4^{2}-4×3×(-1)=16+12=28＞0.\therefore x=\frac {-4\pm \sqrt {28}}{6}=\frac {-4\pm 2\sqrt {7}}{6}=\frac {-2\pm \sqrt {7}}{3}.\therefore x_{1}=\frac {-2+\sqrt {7}}{3},x_{2}=\frac {-2-\sqrt {7}}{3}.$