《2025年名校课堂九年级数学上册湘教版湖南专版》

1. 根据平方根的意义解下列方程:
(1)$4x^{2}-36=0$;
(2)$2(3x - 1)^{2}=8$.

答案： 1.解:
(1)$4x^{2}=36$,$x^{2}=9$,$x=\pm 3$,$\therefore x_{1}=3$,$x_{2}=-3$.
(2)$(3x-1)^{2}=4$,$3x-1=\pm 2$,$\therefore x_{1}=1$,$x_{2}=-\dfrac{1}{3}$.

2. 用配方法解下列方程:
(1)(2024·徐州)$x^{2}+2x - 1=0$;
(2)$\frac{1}{4}x^{2}-6x + 3=0$.

答案： 2.解:
(1)$x^{2}+2x=1$,$x^{2}+2x+1=1+1$,$(x+1)^{2}=2$,$x+1=\pm \sqrt{2}$,$\therefore x_{1}=\sqrt{2}-1$,$x_{2}=-\sqrt{2}-1$.
(2)$x^{2}-24x+12=0$,$(x-12)^{2}=132$,$x-12=\pm 2\sqrt{33}$,$\therefore x_{1}=2\sqrt{33}+12$,$x_{2}=-2\sqrt{33}+12$.

3. 用因式分解法解下列方程:
(1)$2(t - 1)^{2}+8t=0$;
(2)$x^{2}-1=2(x + 1)$;
(3)$2x^{2}-7x + 3=0$.

答案： 3.解:
(1)原方程可化为$2t^{2}+4t+2=0$.$\therefore t^{2}+2t+1=0$.$\therefore (t+1)^{2}=0$.$\therefore t_{1}=t_{2}=-1$.
(2)$(x+1)(x-1)=2(x+1)$,$(x+1)(x-1)-2(x+1)=0$,$(x+1)(x-3)=0$,$\therefore x+1=0$或$x-3=0$.解得$x_{1}=-1$,$x_{2}=3$.
(3)$(2x-1)(x-3)=0$,$\therefore 2x-1=0$或$x-3=0$,解得$x_{1}=\dfrac{1}{2}$,$x_{2}=3$.

4. 用公式法解下列方程:
(1)$9x^{2}-6x + 1=0$;
(2)$5x^{2}-3x=x + 1$;
(3)$x^{2}-2\sqrt{3}x + 2=0$.

答案： 4.解:
(1)这里$a=9$,$b=-6$,$c=1$.因而$b^{2}-4ac=(-6)^{2}-4× 9× 1=0$.所以$x=\dfrac{6\pm \sqrt{0}}{2}$.因此,原方程的根为$x_{1}=x_{2}=\dfrac{1}{3}$.
(2)方程化为一般形式为$5x^{2}-4x-1=0$,$\because a=5$,$b=-4$,$c=-1$,$\therefore b^{2}-4ac=(-4)^{2}-4× 5× (-1)=36＞0$.$\therefore x=\dfrac{4\pm \sqrt{36}}{2× 5}=\dfrac{4\pm 6}{10}$.$\therefore x_{1}=-\dfrac{1}{5}$,$x_{2}=1$.
(3)$\because a=1$,$b=-2\sqrt{3}$,$c=2$,$b^{2}-4ac=(-2\sqrt{3})^{2}-4× 1× 2=4＞0$,$\therefore x=\dfrac{-(-2\sqrt{3})\pm 2}{2× 1}=\sqrt{3}\pm 1$.$\therefore x_{1}=\sqrt{3}-1$,$x_{2}=\sqrt{3}+1$.