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電子課本網 › 第6頁

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信息發(fā)布者：
16
4
$\frac{1}{64}$
$\frac{1}{8}$
±7
D
B
解：移項，得$x^2 + 2x = 1$配方，得$x^2 + 2x + 1 = 1 + 1，$即$(x + 1)^2 = 2$開平方，得$x + 1 = \pm\sqrt{2}$解得$x_1 = -1 + \sqrt{2}，$$x_2 = -1 - \sqrt{2}$
解：移項，得$x^2 - 4x = 3$配方，得$x^2 - 4x + 4 = 3 + 4，$即$(x - 2)^2 = 7$開平方，得$x - 2 = \pm\sqrt{7}$解得$x_1 = 2 + \sqrt{7}，$$x_2 = 2 - \sqrt{7}$
解：移項，得$y^2 - 6y = -6$配方，得$y^2 - 6y + 9 = -6 + 9，$即$(y - 3)^2 = 3$開平方，得$y - 3 = \pm\sqrt{3}$解得$y_1 = 3 + \sqrt{3}，$$y_2 = 3 - \sqrt{3}$
解：移項，得$x^2 - 2\sqrt{3}x = -3$配方，得$x^2 - 2\sqrt{3}x + 3 = -3 + 3，$即$(x - \sqrt{3})^2 = 0$開平方，得$x - \sqrt{3} = 0$解得$x_1 = x_2 = \sqrt{3}$
解：$x^2 - \frac{5}{3}x = 1$
$x^2 - \frac{5}{3}x + (\frac{5}{6})^2 = 1 + (\frac{5}{6})^2$
$(x - \frac{5}{6})^2 = \frac{61}{36}$
$x - \frac{5}{6} = \pm \frac{\sqrt{61}}{6}$
$x_1 = \frac{5 + \sqrt{61}}{6}，$$x_2 = \frac{5 - \sqrt{61}}{6}$
解：$y^2 - \frac{1}{2}y = -\frac{3}{2}$
$y^2 - \frac{1}{2}y + (\frac{1}{4})^2 = -\frac{3}{2} + (\frac{1}{4})^2$
$(y - \frac{1}{4})^2 = -\frac{23}{16}$
此方程無實數根
解：$x^2 + 4x = -2$
$x^2 + 4x + 4 = -2 + 4$
$(x + 2)^2 = 2$
$x + 2 = \pm \sqrt{2}$
$x_1 = -2 + \sqrt{2}，$$x_2 = -2 - \sqrt{2}$
$ 解：x^2 + 9x + 8 = -12 x^2 + 9x = -20 x^2 + 9x + (\frac{9}{2})^2 = -20 + (\frac{9}{2})^2 (x + \frac{9}{2})^2 = \frac{1}{4} x + \frac{9}{2} = \pm \frac{1}{2} x_1 = -4，x_2 = -5 $
解：①當$x \geqslant 1$時，原方程可化為$x^2-(x - 1)-1=0，$即$x^2 - x=0。$解得$x_1=0$（不合題意，舍去），$x_2=1。$②當$x ＜ 1$時，原方程可化為$x^2+(x - 1)-1=0，$即$x^2 + x - 2=0。$解得$x_1=-2，$$x_2=1$（不合題意，舍去）。$\therefore$原方程的根是$x_1=-2，$$x_2=1。$
（1）解：將$x=1$代入方程$x^2 + 2(2 - m)x + 3 - 6m = 0，$得$1 + 2(2 - m) + 3 - 6m = 0，$化簡得$1 + 4 - 2m + 3 - 6m = 0，$$8 - 8m = 0，$解得$m=1。$原方程為$x^2 + 2(2 - 1)x + 3 - 6×1 = x^2 + 2x - 3 = 0，$因式分解得$(x + 3)(x - 1) = 0，$方程的另一個根為$x=-3。$
（2）證明：$\Delta = [2(2 - m)]^2 - 4×1×(3 - 6m) = 4(4 - 4m + m^2) - 12 + 24m = 16 - 16m + 4m^2 - 12 + 24m = 4m^2 + 8m + 4 = 4(m + 1)^2。$因為$(m + 1)^2 \geq 0，$所以$\Delta = 4(m + 1)^2 \geq 0，$無論$m$取何值，此方程總有實數根。
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