《2025年名师帮同步学案九年级数学全一册人教版》

一、新课学习
A. 形如$x^{2}= p(p≥0)$,直接开平方法可得$ x = \pm \sqrt{p} $.
B. 形如$ax^{2}= p(a＞0,p≥0)$,则$ x = \pm \sqrt{\frac{p}{a}} $.
C. 形如$(mx+n)^{2}= p(p≥0)$,则$mx+n= $$ \pm \sqrt{p} $.

答案： A. $ x = \pm \sqrt{p} $ B. $ x = \pm \sqrt{\frac{p}{a}} $ C. $ \pm \sqrt{p} $

1. 方程$x^{2}= 16$的解为$ x = \pm 4 $.

答案： $ x = \pm 4 $

2. 方程$4x^{2}= 1$的解为$ x = \pm \frac{1}{2} $.

答案： $ x = \pm \frac{1}{2} $

3. 方程$(2x - 3)^{2}= 9$的解为$ x_1 = 3, x_2 = 0 $.

答案： $ x_1 = 3, x_2 = 0 $

【例题1】用直接开方法解方程.
(1)$x^{2}= 25$；
解:将方程两边开方,
解得 $ x = \pm \sqrt{25} $,
$ \therefore x_1 = $5$, x_2 = $-5;
(2)$x^{2}-12= 0$.
解: $ x^2 = $12,
将方程两边开方,
解得 $ x = \pm \sqrt{12} $,
$ \therefore x_1 = $$2\sqrt{3}$$, x_2 = $$-2\sqrt{3}$.

答案：
(1)解:将方程两边开方,
解得 $ x = \pm \sqrt{25} $,
$ \therefore x_1 = 5, x_2 = -5 $;
(2)解: $ x^2 = 12 $,
将方程两边开方,
解得 $ x = \pm \sqrt{12} $,
$ \therefore x_1 = 2\sqrt{3}, x_2 = -2\sqrt{3} $.

【变式1】用直接开方法解方程.
(1)$x^{2}= 3$；
解:将方程两边开方,
解得 $ x = $$\pm \sqrt{3}$,
$ \therefore x_1 = $$\sqrt{3}$, $ x_2 = $$-\sqrt{3}$;
(2)$y^{2}-9= 0$.
解: $ y^2 = $9,
将方程两边开方,
解得 $ y = $$\pm \sqrt{9}$,
$ \therefore y_1 = $3, $ y_2 = $-3.

答案：
(1)解:将方程两边开方,
解得 $ x = \pm \sqrt{3} $,
$ \therefore x_1 = \sqrt{3}, x_2 = -\sqrt{3} $;
(2)解: $ y^2 = 9 $,
将方程两边开方,
解得 $ y = \pm \sqrt{9} $,
$ \therefore y_1 = 3, y_2 = -3 $.

【例题2】用直接开方法解方程.
(1)$9x^{2}= 4$；
解: $ x^2 = $$\frac{4}{9}$,
将方程两边开方,
解得 $ x = \pm \sqrt{}$$\frac{4}{9}$,
$ \therefore x_1 = $$\frac{2}{3}$, $ x_2 = $$-\frac{2}{3}$;
(2)$16x^{2}-49= 0$.
解: $ x^2 = $$\frac{49}{16}$,
将方程两边开方,
解得 $ x = \pm \sqrt{}$$\frac{49}{16}$,
$ \therefore x_1 = $$\frac{7}{4}$, $ x_2 = $$-\frac{7}{4}$.

答案：
(1)解: $ x^2 = \frac{4}{9} $,
将方程两边开方,
解得 $ x = \pm \sqrt{\frac{4}{9}} $,
$ \therefore x_1 = \frac{2}{3}, x_2 = -\frac{2}{3} $;
(2)解: $ x^2 = \frac{49}{16} $,
将方程两边开方,
解得 $ x = \pm \sqrt{\frac{49}{16}} $,
$ \therefore x_1 = \frac{7}{4}, x_2 = -\frac{7}{4} $.

【变式2】用直接开方法解方程.
(1)$25x^{2}= 9$；
解: $ x^2 = $$\frac{9}{25}$,
将方程两边开方,
解得 $ x = \pm $$\sqrt{\frac{9}{25}}$,
$ \therefore x_1 = $$\frac{3}{5}$, $ x_2 = $$-\frac{3}{5}$;
(2)$6x^{2}-12= 0$.
解: $ 6x^2 = $12, $ x^2 = $2,
将方程两边开方,
解得 $ x = \pm $$\sqrt{2}$,
$ \therefore x_1 = $$\sqrt{2}$, $ x_2 = $$-\sqrt{2}$.

答案：
(1)解: $ x^2 = \frac{9}{25} $,
将方程两边开方,
解得 $ x = \pm \sqrt{\frac{9}{25}} $,
$ \therefore x_1 = \frac{3}{5}, x_2 = -\frac{3}{5} $;
(2)解: $ 6x^2 = 12, x^2 = 2 $,
将方程两边开方,
解得 $ x = \pm \sqrt{2} $,
$ \therefore x_1 = \sqrt{2}, x_2 = -\sqrt{2} $.

【例题3】用直接开方法解方程.
(1)$(x + 2)^{2}= 9$；
解:将方程两边开方,得
$ x + 2 = \pm 3 $,
$ \therefore x_1 = $1, $ x_2 = $-5;
(2)$2(x - 4)^{2}= 50$.
解:整理得 $ (x - 4)^2 = $25,
将方程两边开方,
得 $ x - 4 = \pm 5 $,
$ \therefore x_1 = $9, $ x_2 = $-1.

答案：
(1)解:将方程两边开方,得
$ x + 2 = \pm 3 $,
$ \therefore x_1 = 1, x_2 = -5 $;
(2)解:整理得 $ (x - 4)^2 = 25 $,
将方程两边开方,
得 $ x - 4 = \pm 5 $,
$ \therefore x_1 = 9, x_2 = -1 $.

【变式3】用直接开方法解方程.
(1)$100(1 + x)^{2}= 121$；
解:将方程变形为
$ (1 + x)^2 = \frac{121}{100} $,
两边开方,
得 $ 1 + x = \pm \frac{11}{10} $,
$ \therefore x_1 = $$\frac{1}{10}$, $ x_2 = $$-\frac{21}{10}$；
(2)$81(1 - x)^{2}= 64$.
解:将方程变形为 $ (1 - x)^2 = \frac{64}{81} $,
两边开方,
得 $ 1 - x = \pm \frac{8}{9} $,
$ \therefore x_1 = $$\frac{1}{9}$, $ x_2 = $$\frac{17}{9}$.

答案：
(1)解:将方程变形为
$ (1 + x)^2 = \frac{121}{100} $,
两边开方,
得 $ 1 + x = \pm \frac{11}{10} $,
$ \therefore x_1 = \frac{1}{10}, x_2 = -\frac{21}{10} $;
(2)解:将方程变形为 $ (1 - x)^2 = \frac{64}{81} $,
两边开方,
得 $ 1 - x = \pm \frac{8}{9} $,
$ \therefore x_1 = \frac{1}{9}, x_2 = \frac{17}{9} $.