《2025年名校课堂九年级数学上册人教版安徽专版》

1. 用直接开平方法解下列方程：
(1)$(2x + 1)^2 = 9$.
(2)$3(x + 1)^2 - 108 = 0$.

答案： 1.解:
(1)$2x+1=\pm 3$.$\therefore 2x+1=3$或$2x+1=-3$.$\therefore x_{1}=1$,$x_{2}=-2$.
(2)$3(x+1)^{2}=108$,$(x+1)^{2}=36$.$\therefore x+1=\pm 6$.$\therefore x_{1}=5$,$x_{2}=-7$.

2. 用配方法解下列方程：
(1)$x^2 - 2x - 99 = 0$.
(2)$x^2 + 12x = -27$.

答案： 2.解:
(1)$x^{2}-2x=99$,$x^{2}-2x+1=99+1$,即$(x-1)^{2}=100$.$\therefore x-1= \pm 10$.$\therefore x_{1}=11$,$x_{2}=-9$.
(2)$x^{2}+12x+36=-27+36$,即$(x+6)^{2}=9$.$\therefore x+6=\pm 3$.$\therefore x_{1}=-3$,$x_{2}=-9$.

3. 用因式分解法解下列方程：
(1)$x(2x + 1) = 3(2x + 1)$.
(2)$9x^2 - (x - 1)^2 = 0$.

答案： 3.解:
(1)$x(2x+1)-3(2x+1)=0$,$(2x+1)(x-3)=0$.$\therefore 2x+1=0$或$x-3=0$.$\therefore x_{1}=-\dfrac{1}{2}$,$x_{2}=3$.
(2)$(3x+x-1)(3x-x+1)=0$,$(4x-1)(2x+1)=0$,$\therefore 4x-1=0$或$2x+1=0$.$\therefore x_{1}=\dfrac{1}{4}$,$x_{2}=-\dfrac{1}{2}$.

4. 用公式法解下列方程：
(1)$2x^2 + 2x - 1 = 0$.
(2)$4x^2 - 12x + 9 = 0$.

答案： 4.解:
(1)$\because a=2$,$b=2$,$c=-1$,$\therefore \Delta =b^{2}-4ac=2^{2}-4× 2×(-1)=12＞0$.$\therefore x=\dfrac{-2\pm \sqrt{12}}{2× 2}=\dfrac{-2\pm 2\sqrt{3}}{4}=\dfrac{-1\pm \sqrt{3}}{2}$.$\therefore x_{1}=\dfrac{-1+\sqrt{3}}{2}$,$x_{2}=\dfrac{-1-\sqrt{3}}{2}$.
(2)$\because a=4$,$b=-12$,$c=9$,$\therefore \Delta =b^{2}-4ac=(-12)^{2}-4× 4× 9=0$.$\therefore x=\dfrac{12\pm \sqrt{0}}{2× 4}=\dfrac{12}{8}$.$\therefore x_{1}=x_{2}=\dfrac{3}{2}$.