《2025年启东中学作业本九年级数学上册苏科版宿迁专版》

用适当的方法解一元二次方程：
1. $ 4(x - 1)^2 = 9 $.
2. $ x^2 + 4x - 96 = 0 $.
3. $ 2x^2 - 3x - 5 = 0 $.
4. $ 2x^2 - 7x + 4 = 1 $.
5. $ \sqrt{3}x^2 + 6\sqrt{3}x + 9\sqrt{3} = 0 $.
6. $ x(x + 3) = 2x + 6 $.
7. $ 3(x - 5)^2 = 2(5 - x) $.
8. $ 2(x - 3)^2 = x^2 - 9 $.
9. $ (x - 1)(x + 3) = 12 $.
10. $ (x + 5)^2 - 4(x + 5) = 5 $.

答案： 1. 解：
$(x - 1)^2=\frac{9}{4}$
$x - 1=\pm\frac{3}{2}$
$x = 1\pm\frac{3}{2}$
$x_1=\frac{5}{2},x_2=-\frac{1}{2}$
2. 解：
$x^2 + 4x - 96 = 0$
$(x + 12)(x - 8)=0$
$x + 12 = 0$或$x - 8 = 0$
$x_1=-12,x_2=8$
3. 解：
$2x^2 - 3x - 5 = 0$
$(2x - 5)(x + 1)=0$
$2x - 5 = 0$或$x + 1 = 0$
$x_1=\frac{5}{2},x_2=-1$
4. 解：
$2x^2 - 7x + 4 = 1$
$2x^2 - 7x + 3 = 0$
$(2x - 1)(x - 3)=0$
$2x - 1 = 0$或$x - 3 = 0$
$x_1=\frac{1}{2},x_2=3$
5. 解：
$\sqrt{3}x^2 + 6\sqrt{3}x + 9\sqrt{3} = 0$
$x^2 + 6x + 9 = 0$
$(x + 3)^2 = 0$
$x_1=x_2=-3$
6. 解：
$x(x + 3) = 2x + 6$
$x(x + 3) - 2(x + 3)=0$
$(x + 3)(x - 2)=0$
$x + 3 = 0$或$x - 2 = 0$
$x_1=-3,x_2=2$
7. 解：
$3(x - 5)^2 = 2(5 - x)$
$3(x - 5)^2 + 2(x - 5)=0$
$(x - 5)(3x - 15 + 2)=0$
$(x - 5)(3x - 13)=0$
$x - 5 = 0$或$3x - 13 = 0$
$x_1=5,x_2=\frac{13}{3}$
8. 解：
$2(x - 3)^2 = x^2 - 9$
$2(x - 3)^2 - (x + 3)(x - 3)=0$
$(x - 3)(2x - 6 - x - 3)=0$
$(x - 3)(x - 9)=0$
$x - 3 = 0$或$x - 9 = 0$
$x_1=3,x_2=9$
9. 解：
$(x - 1)(x + 3) = 12$
$x^2 + 2x - 3 - 12 = 0$
$x^2 + 2x - 15 = 0$
$(x + 5)(x - 3)=0$
$x + 5 = 0$或$x - 3 = 0$
$x_1=-5,x_2=3$
10. 解：
$(x + 5)^2 - 4(x + 5) = 5$
$(x + 5)^2 - 4(x + 5) - 5 = 0$
$(x + 5 - 5)(x + 5 + 1)=0$
$x(x + 6)=0$
$x = 0$或$x + 6 = 0$
$x_1=0,x_2=-6$