1.用直接开平方法解方程：
(1)$3(2x-\sqrt {2})^{2}= 54$.
$x_1 = 2\sqrt{2}$，$x_2 = -\sqrt{2}$
(2)$(x-1)^{2}= (2x+3)^{2}$.
$x_1 = -4$，$x_2 = -\frac{2}{3}$
答案：解析
(1) $ 3(2x - \sqrt{2})^2 = 54 $，整理，得 $ (2x - \sqrt{2})^2 = 18 $，直接开平方，得 $ 2x - \sqrt{2} = \pm 3\sqrt{2} $，解得 $ x_1 = 2\sqrt{2} $，$ x_2 = -\sqrt{2} $。
(2) $ (x - 1)^2 = (2x + 3)^2 $，直接开平方，得 $ x - 1 = \pm (2x + 3) $，即 $ x - 1 = 2x + 3 $ 或 $ x - 1 = -2x - 3 $，解得 $ x_1 = -4 $，$ x_2 = -\frac{2}{3} $。
2.用配方法解方程：
(1)$x^{2}-4x+1= 0$.
$ x_1 = 2 + \sqrt{3} $，$ x_2 = 2 - \sqrt{3} $
(2)$3x^{2}-8x-3= 0$.
$ x_1 = 3 $，$ x_2 = -\frac{1}{3} $
(3)$-x^{2}+4x= x^{2}-3$.
$ x_1 = \frac{2 + \sqrt{10}}{2} $，$ x_2 = \frac{2 - \sqrt{10}}{2} $
(4)$x^{2}-2\sqrt {2}x-4= 0$.
$ x_1 = \sqrt{2} + \sqrt{6} $，$ x_2 = \sqrt{2} - \sqrt{6} $
答案：解析
(1) $ x^2 - 4x + 1 = 0 $，移项，得 $ x^2 - 4x = -1 $，配方，得 $ x^2 - 4x + 4 = -1 + 4 $，即 $ (x - 2)^2 = 3 $，$ \therefore x - 2 = \pm \sqrt{3} $，解得 $ x_1 = 2 + \sqrt{3} $，$ x_2 = 2 - \sqrt{3} $。
(2) $ 3x^2 - 8x - 3 = 0 $，移项，得 $ 3x^2 - 8x = 3 $，二次项系数化为1，得 $ x^2 - \frac{8}{3}x = 1 $，配方，得 $ x^2 - \frac{8}{3}x + (-\frac{4}{3})^2 = 1 + (-\frac{4}{3})^2 $，即 $ (x - \frac{4}{3})^2 = \frac{25}{9} $，直接开平方，得 $ x - \frac{4}{3} = \pm \frac{5}{3} $，解得 $ x_1 = 3 $，$ x_2 = -\frac{1}{3} $。
(3) 移项，得 $ -2x^2 + 4x = -3 $，二次项系数化为1，得 $ x^2 - 2x = \frac{3}{2} $，配方，得 $ x^2 - 2x + 1 = \frac{3}{2} + 1 $，即 $ (x - 1)^2 = \frac{5}{2} $，解得 $ x_1 = \frac{2 + \sqrt{10}}{2} $，$ x_2 = \frac{2 - \sqrt{10}}{2} $。
(4) 移项，得 $ x^2 - 2\sqrt{2}x = 4 $，配方，得 $ x^2 - 2\sqrt{2}x + 2 = 4 + 2 $，即 $ (x - \sqrt{2})^2 = 6 $，解得 $ x_1 = \sqrt{2} + \sqrt{6} $，$ x_2 = \sqrt{2} - \sqrt{6} $。
3.用公式法解下列方程：
(1)$2x^{2}-9x+10= 0$.
$x_1 = \frac{5}{2}$，$x_2 = 2$
(2)$7x^{2}+2x+3= 0$.
原方程无实数根
(3)$4x^{2}+4x= -1$.
$x_1 = x_2 = -\frac{1}{2}$
(4)$x^{2}-\sqrt {3}x-\frac {1}{4}= 0$.
$x_1 = \frac{\sqrt{3} + 2}{2}$，$x_2 = \frac{\sqrt{3} - 2}{2}$
答案：解析
(1) $ 2x^2 - 9x + 10 = 0 $，$ \because a = 2 $，$ b = -9 $，$ c = 10 $，$ \therefore \Delta = b^2 - 4ac = (-9)^2 - 4 \times 2 \times 10 = 81 - 80 = 1 ＞ 0 $，$ \therefore $ 方程有两个不相等的实数根，$ \therefore x = \frac{-(-9) \pm \sqrt{1}}{2 \times 2} = \frac{9 \pm 1}{4} $，$ \therefore x_1 = \frac{5}{2} $，$ x_2 = 2 $。
(2) $ 7x^2 + 2x + 3 = 0 $，$ \because a = 7 $，$ b = 2 $，$ c = 3 $，$ \therefore \Delta = b^2 - 4ac = 2^2 - 4 \times 7 \times 3 = 4 - 84 = -80 ＜ 0 $，$ \therefore $ 原方程无实数根。
(3) $ 4x^2 + 4x = -1 $，整理，得 $ 4x^2 + 4x + 1 = 0 $，$ \because a = 4 $，$ b = 4 $，$ c = 1 $，$ \therefore \Delta = b^2 - 4ac = 4^2 - 4 \times 4 \times 1 = 0 $，$ \therefore $ 方程有两个相等的实数根，$ \therefore x_1 = x_2 = -\frac{b}{2a} = -\frac{4}{2 \times 4} = -\frac{1}{2} $。
(4) $ x^2 - \sqrt{3}x - \frac{1}{4} = 0 $，$ \because a = 1 $，$ b = -\sqrt{3} $，$ c = -\frac{1}{4} $，$ \therefore \Delta = b^2 - 4ac = (-\sqrt{3})^2 - 4 \times 1 \times (-\frac{1}{4}) = 4 ＞ 0 $，$ \therefore $ 方程有两个不相等的实数根，$ \therefore x = \frac{\sqrt{3} \pm \sqrt{4}}{2 \times 1} = \frac{\sqrt{3} \pm 2}{2} $，$ \therefore x_1 = \frac{\sqrt{3} + 2}{2} $，$ x_2 = \frac{\sqrt{3} - 2}{2} $。
4.用因式分解法解下列方程：
(1)$(x+2)^{2}= 8+4x$.
$x_1 = -2$，$ x_2 = 2$
(2)$5(x-3)^{2}= x^{2}-9$.
$x_1 = 3$，$ x_2 = \frac{9}{2}$
(3)$4(x+3)^{2}-9(x-3)^{2}= 0$.
$x_1 = \frac{3}{5}$，$ x_2 = 15$
(4)$x^{2}-2x-15= 0$.
$x_1 = -3$，$ x_2 = 5$
答案：解析
(1) $ \because (x + 2)^2 = 8 + 4x $，$ \therefore (x + 2)^2 - 4x - 8 = 0 $，$ \therefore (x + 2)^2 - 4(x + 2) = 0 $，因式分解，得 $ (x + 2)(x + 2 - 4) = 0 $，即 $ (x + 2)(x - 2) = 0 $，$ \therefore x + 2 = 0 $ 或 $ x - 2 = 0 $，解得 $ x_1 = -2 $，$ x_2 = 2 $。
(2) $ 5(x - 3)^2 = x^2 - 9 $，移项，得 $ 5(x - 3)^2 - (x^2 - 9) = 0 $，$ \therefore 5(x - 3)^2 - (x + 3)(x - 3) = 0 $，因式分解，得 $ (x - 3)(5x - 15 - x - 3) = 0 $，即 $ (x - 3)(4x - 18) = 0 $，则 $ x - 3 = 0 $ 或 $ 4x - 18 = 0 $，解得 $ x_1 = 3 $，$ x_2 = \frac{9}{2} $。
(3) 因式分解，得 $ [2(x + 3) + 3(x - 3)][2(x + 3) - 3(x - 3)] = 0 $，即 $ (5x - 3)(-x + 15) = 0 $，解得 $ x_1 = \frac{3}{5} $，$ x_2 = 15 $。
(4) 因式分解，得 $ (x + 3)(x - 5) = 0 $，$ \therefore x + 3 = 0 $ 或 $ x - 5 = 0 $，$ \therefore x_1 = -3 $，$ x_2 = 5 $。
5.已知关于x的一元二次方程$x^{2}-x-6= 0的解为x_{1}= -2,x_{2}= 3$,则$(2x+3)^{2}-(2x+3)-6= 0$的解为
$ x_1 = -\frac{5}{2} $，$ x_2 = 0 $
.
答案：答案 $ x_1 = -\frac{5}{2} $，$ x_2 = 0 $
解析 $ \because x^2 - x - 6 = 0 $ 的解为 $ x_1 = -2 $，$ x_2 = 3 $，$ \therefore (2x + 3)^2 - (2x + 3) - 6 = 0 $ 中 $ 2x + 3 = -2 $ 或 $ 3 $，解得 $ x_1 = -\frac{5}{2} $，$ x_2 = 0 $。
6.已知方程$(x^{2}+y^{2})^{2}-(x^{2}+y^{2})-6= 0$,则$x^{2}+y^{2}=$
3
.
答案：答案 3
解析 设 $ x^2 + y^2 = t $，则原方程可化为 $ t^2 - t - 6 = 0 $，解得 $ t_1 = -2 $，$ t_2 = 3 $，又 $ \because x^2 + y^2 \geq 0 $，$ \therefore x^2 + y^2 = t = 3 $。
7.已知x为实数,且满足$(x^{2}+x+1)^{2}+2(x^{2}+x+1)-3= 0$,那么$x^{2}+x+1$的值为______
1
。
答案：答案 1
解析 设 $ x^2 + x + 1 = t $，则原方程可化为 $ t^2 + 2t - 3 = 0 $，解得 $ t_1 = 1 $，$ t_2 = -3 $，又 $ \because x^2 + x + 1 = (x^2 + x + \frac{1}{4}) + 1 - \frac{1}{4} = (x + \frac{1}{2})^2 + \frac{3}{4} \geq \frac{3}{4} $，$ \therefore x^2 + x + 1 = 1 $。
