《2025年实验班提优训练九年级数学上册人教版》

1. 解方程:
(1)$2x^{2}-8= 0$;$\because 2 x^{2}-8=0, \therefore x^{2}=4, \therefore x_{1}=2, x_{2}=-2$
(2)$(x+5)^{2}= 25$;方程开方, 得 $x+5=5$ 或 $x+5=-5$,解得 $x_{1}=0, x_{2}=-10$
(3)(2024·四川内江期中)$4(x-1)^{2}-36= 0$;$\because 4(x-1)^{2}-36=0, \therefore(x-1)^{2}=9,\therefore x-1=\pm 3, \therefore x_{1}=4, x_{2}=-2$
(4)$(3x-1)^{2}= 4(2x+3)^{2}$.$\because(3 x-1)^{2}=4(2 x+3)^{2}, \therefore 3 x-1=\pm 2(2 x+3),\therefore 3 x-1=4 x+6$ 或 $3 x-1=-4 x-6,\therefore x_{1}=-7, x_{2}=-\frac{5}{7}$

答案：
(1) $\because 2 x^{2}-8=0, \therefore x^{2}=4, \therefore x_{1}=2, x_{2}=-2$.
(2) 方程开方, 得 $x+5=5$ 或 $x+5=-5$,
解得 $x_{1}=0, x_{2}=-10$.
(3) $\because 4(x-1)^{2}-36=0, \therefore(x-1)^{2}=9$,
$\therefore x-1=\pm 3, \therefore x_{1}=4, x_{2}=-2$.
(4) $\because(3 x-1)^{2}=4(2 x+3)^{2}, \therefore 3 x-1=\pm 2(2 x+3)$,
$\therefore 3 x-1=4 x+6$ 或 $3 x-1=-4 x-6$,
$\therefore x_{1}=-7, x_{2}=-\frac{5}{7}$.

2. 解方程:
(1)(2025·河南驻马店驿城区期中)$x^{2}-4x= 6;$
配方, 得 $x^{2}-4 x+4=6+4$,即 $(x-2)^{2}=10$, 开方, 得 $x-2=\pm \sqrt{10}$,解得 $x_{1}=2+\sqrt{10}, x_{2}=2-\sqrt{10}$.
(2)$2x^{2}-3x= -1;$
原方程整理, 得 $x^{2}-\frac{3}{2} x=-\frac{1}{2}$,配方, 得 $x^{2}-\frac{3}{2} x+\frac{9}{16}=\frac{1}{16}$, 即 $\left(x-\frac{3}{4}\right)^{2}=\frac{1}{16}$,开方, 得 $x=\frac{3}{4} \pm \frac{1}{4}$, 解得 $x_{1}=1, x_{2}=\frac{1}{2}$.
(3)$x^{2}-2x-1= 0;$
移项, 得 $x^{2}-2 x=1$,配方, 得 $x^{2}-2 x+1=2$, 即 $(x-1)^{2}=2$,开方, 得 $x-1=\pm \sqrt{2}$, 解得 $x_{1}=1+\sqrt{2}, x_{2}=1-\sqrt{2}$.
(4)$2x^{2}-8x= -7.$
整理, 得 $x^{2}-4 x=-\frac{7}{2}$,配方, 得 $x^{2}-4 x+4=-\frac{7}{2}+4$, 即 $(x-2)^{2}=\frac{1}{2}$,开方, 得 $x-2=\pm \frac{\sqrt{2}}{2}$, 解得 $x_{1}=2+\frac{\sqrt{2}}{2}, x_{2}=2-\frac{\sqrt{2}}{2}$.

答案：
(1) 配方, 得 $x^{2}-4 x+4=6+4$,
即 $(x-2)^{2}=10$, 开方, 得 $x-2=\pm \sqrt{10}$,
解得 $x_{1}=2+\sqrt{10}, x_{2}=2-\sqrt{10}$.
(2) 原方程整理, 得 $x^{2}-\frac{3}{2} x=-\frac{1}{2}$,
配方, 得 $x^{2}-\frac{3}{2} x+\frac{9}{16}=\frac{1}{16}$, 即 $\left(x-\frac{3}{4}\right)^{2}=\frac{1}{16}$,
开方, 得 $x=\frac{3}{4} \pm \frac{1}{4}$, 解得 $x_{1}=1, x_{2}=\frac{1}{2}$.
(3) 移项, 得 $x^{2}-2 x=1$,
配方, 得 $x^{2}-2 x+1=2$, 即 $(x-1)^{2}=2$,
开方, 得 $x-1=\pm \sqrt{2}$, 解得 $x_{1}=1+\sqrt{2}, x_{2}=1-\sqrt{2}$.
(4) 整理, 得 $x^{2}-4 x=-\frac{7}{2}$,
配方, 得 $x^{2}-4 x+4=-\frac{7}{2}+4$, 即 $(x-2)^{2}=\frac{1}{2}$,
开方, 得 $x-2=\pm \frac{\sqrt{2}}{2}$, 解得 $x_{1}=2+\frac{\sqrt{2}}{2}, x_{2}=2-\frac{\sqrt{2}}{2}$.

3. 解方程:
(1)(2025·河南驻马店十中期中)$3x^{2}-2x-2= 0;$
$\because a=3, b=-2, c=-2$,
$\therefore b^{2}-4 a c=(-2)^{2}-4 × 3 ×(-2)=28＞0$,
$\therefore x=\frac{-(-2) \pm \sqrt{28}}{6}=\frac{1 \pm \sqrt{7}}{3}, \therefore x_{1}=\frac{1+\sqrt{7}}{3}, x_{2}=\frac{1-\sqrt{7}}{3}$.
(2)$\frac {1}{2}x^{2}-5x+4= 0;$
$\because a=\frac{1}{2}, b=-5, c=4, \therefore \Delta=b^{2}-4 a c=25-8=17＞0$,
$\therefore x=\frac{5 \pm \sqrt{17}}{2 × \frac{1}{2}}=5 \pm \sqrt{17}, \therefore x_{1}=5+\sqrt{17}, x_{2}=5-\sqrt{17}$.
(3)(2025·四川成都期中)$2x^{2}-3(x+1)= 0;$
整理, 得 $2 x^{2}-3 x-3=0 . \because a=2, b=-3, c=-3$,
$\therefore \Delta=(-3)^{2}-4 × 2 ×(-3)=33＞0$,
则 $x=\frac{3 \pm \sqrt{33}}{4}$, 即 $x_{1}=\frac{3+\sqrt{33}}{4}, x_{2}=\frac{3-\sqrt{33}}{4}$.
(4)$(x+3)(x-1)= 3.$
整理成一般式, 得 $x^{2}+2 x-6=0$,
$\because a=1, b=2, c=-6, \therefore \Delta=2^{2}-4 × 1 ×(-6)=28＞0$,
$\therefore x=\frac{-2 \pm 2 \sqrt{7}}{2 × 1}, \therefore x_{1}=-1+\sqrt{7}, x_{2}=-1-\sqrt{7}$.

答案：
(1) $\because a=3, b=-2, c=-2$,
$\therefore b^{2}-4 a c=(-2)^{2}-4 \times 3 \times(-2)=28＞0$,
$\therefore x=\frac{-(-2) \pm \sqrt{28}}{6}=\frac{1 \pm \sqrt{7}}{3}, \therefore x_{1}=\frac{1+\sqrt{7}}{3}, x_{2}=\frac{1-\sqrt{7}}{3}$.
(2) $\because a=\frac{1}{2}, b=-5, c=4, \therefore \Delta=b^{2}-4 a c=25-8=17＞0$,
$\therefore x=\frac{5 \pm \sqrt{17}}{2 \times \frac{1}{2}}=5 \pm \sqrt{17}, \therefore x_{1}=5+\sqrt{17}, x_{2}=5-\sqrt{17}$.
(3) 整理, 得 $2 x^{2}-3 x-3=0 . \because a=2, b=-3, c=-3$,
$\therefore \Delta=(-3)^{2}-4 \times 2 \times(-3)=33＞0$,
则 $x=\frac{3 \pm \sqrt{33}}{4}$, 即 $x_{1}=\frac{3+\sqrt{33}}{4}, x_{2}=\frac{3-\sqrt{33}}{4}$.
(4) 整理成一般式, 得 $x^{2}+2 x-6=0$,
$\because a=1, b=2, c=-6, \therefore \Delta=2^{2}-4 \times 1 \times(-6)=28＞0$,
$\therefore x=\frac{-2 \pm 2 \sqrt{7}}{2 \times 1}, \therefore x_{1}=-1+\sqrt{7}, x_{2}=-1-\sqrt{7}$.