零五网 › 全部参考答案› 补充习题答案 › 苏科版数学补充习题九年级上下册答案 › 第4页

第4页

信息发布者：

解：$x^2-6x = 7$

$ x^2-6x + 9 = 7 + 9$

$ (x - 3)^2=16$

$ x - 3=\pm 4$

$ x_{1}=7，$$x_{2}=-1$

解：$x^2+10x + 9 = 0$

$ x^2+10x + 25=-9 + 25$

$ (x + 5)^2=16$

$ x + 5=\pm 4$

$x_{1}=-1，$$x_{2}=-9$

解：$x^2+2x = 7$

$ (x + 1)^2=8$

$ x + 1=\pm 2\sqrt {2}$

$ x_{1}=-1 + 2\sqrt {2}，$$x_{2}=-1 - 2\sqrt {2}$

解：$x^2-6x - 1 = 0$

$x^2-6x + 9 = 1 + 9$

$ (x - 3)^2=10$

$ x - 3=\pm \sqrt {10}$

$ x_{1}=3+\sqrt {10}，$$x_{2}=3-\sqrt {10}$

解：$t^2-t = 2$

$ t^2-t+\frac {1}{4}=2+\frac {1}{4}$

$ (t-\frac {1}{2})^2=\frac {9}{4}$

$ t-\frac {1}{2}=\pm \frac {3}{2}$

$ t_{1}=2，$$t_{2}=-1$

解：$x^2-3=-3x$

$ x^2+3x+\frac {9}{4}=3+\frac {9}{4}$

$ (x+\frac {3}{2})^2=\frac {21}{4}$

$ x+\frac {3}{2}=\pm \frac {\sqrt {21}}{2}$

$ x_{1}=\frac {-3 + \sqrt {21}}{2}，$$x_{2}=\frac {-3 - \sqrt {21}}{2}$

解：$2x^2-5x = 0$

$x^2-\frac {5}{2}x=0$

$x^2-\frac {5}{2}x+\frac {25}{16}=\frac {25}{16}$

$(x-\frac {5}{4})^2=\frac {25}{16}$

$x-\frac {5}{4}=±\frac {5}{4}$

$x_{1}=0，$$x_{2}=\frac {5}{2}$

解：$2x^2+3x + 1 = 0$

$x^2+\frac {3}{2}x=-\frac {1}{2}$

$x^2+\frac {3}{2}x+\frac {9}{16}=-\frac {1}{2}+\frac {9}{16}$

$(x+\frac {3}{4})^2=\frac {1}{16}$

$x+\frac {3}{4}=±\frac {1}{4}$

$x_{1}=-\frac {1}{2}，$$x_{2}=-1$

解：$3x^2-x - 2 = 0$

$x^2-\frac {1}{3}x=\frac {2}{3}$

$x^2-\frac {1}{3}x+\frac {1}{36}=\frac {2}{3}+\frac {1}{36}$

$(x-\frac {1}{6})^2=\frac {25}{36}$

$x-\frac {1}{6}=±\frac {5}{6}$

$x_{1}=-\frac {2}{3}，$$x_{2}=1$

解：$4x^2-2x - 1 = 0$

$x^2-\frac {1}{2}x=\frac {1}{4}$

$x^2-\frac {1}{2}x+\frac {1}{16}=\frac {1}{4}+\frac {1}{16}$

$(x-\frac {1}{4})^2=\frac {5}{16}$

$x-\frac {1}{4}=±\frac {\sqrt 5}4$

$x_{1}=\frac {1 + \sqrt {5}}{4}，$$x_{2}=\frac {1 - \sqrt {5}}{4}$