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第13頁(yè)

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0.618

6或-4

$\frac{1 - \sqrt{13}}{2}$

解：對(duì)于方程?$y^2+2\sqrt {2}y = 6，$?化為一般式為

?$y^2+2\sqrt {2}y - 6 = 0，$?

其中?$a = 1，$??$b = 2\sqrt {2}，$??$c = -6，$?

?$b^2-4ac=(2\sqrt {2})^2-4×1×(-6)=8 + 24 = 32。$?

則?$y=\frac {-2\sqrt {2}\pm \sqrt {32}}{2×1}=\frac {-2\sqrt {2}\pm 4\sqrt {2}}{2}，$?

所以?$y_{1}=\frac {-2\sqrt {2}+4\sqrt {2}}{2}=\sqrt {2}，$??$y_{2}=\frac {-2\sqrt {2}-4\sqrt {2}}{2}=-3\sqrt {2}。$?

解：將方程?$(2x + 1)(x - 1)=8(9 - x)-1$?展開

并化為一般式：?$2x^2-2x+x - 1 = 72 - 8x - 1，$?

即?$2x^2+7x - 72 = 0，$?

其中?$a = 2，$??$b = 7，$??$c = -72，$?

?$b^2-4ac=7^2-4×2×(-72)=49 + 576 = 625。$?

則?$x=\frac {-7\pm \sqrt {625}}{2×2}=\frac {-7\pm 25}{4}，$?

所以?$x_{1}=\frac {-7 + 25}{4}=\frac {9}{2}，$??$x_{2}=\frac {-7 - 25}{4}=-8。$?

解：將方程?$(x + 1)^2-2(x - 1)^2=7$?展開并化為

一般式：?$x^2+2x + 1-2(x^2-2x + 1)=7，$?

?$x^2+2x + 1-2x^2+4x - 2 = 7，$?

即?$-x^2+6x - 8 = 0，$?

兩邊同時(shí)乘以?$-1$?得?$x^2-6x + 8 = 0，$?

其中?$a = 1，$??$b = -6，$??$c = 8，$?

?$b^2-4ac=(-6)^2-4×1×8 = 36 - 32 = 4。$?

則?$x=\frac {-(-6)\pm \sqrt {4}}{2×1}=\frac {6\pm 2}{2}，$?

所以?$x_{1}=\frac {6 + 2}{2}=4，$??$x_{2}=\frac {6 - 2}{2}=2。$?

解：將方程?$1 - t^2=2t(2t - 1)$?展開并化為一般式：

?$1 - t^2=4t^2-2t，$?即?$5t^2-2t - 1 = 0，$?

其中?$a = 5，$??$b = -2，$??$c = -1，$?

?$b^2-4ac=(-2)^2-4×5×(-1)=4 + 20 = 24。$?

則?$t=\frac {-(-2)\pm \sqrt {24}}{2×5}=\frac {2\pm 2\sqrt {6}}{10}=\frac {1\pm \sqrt {6}}{5}，$?

所以?$t_{1}=\frac {1+\sqrt {6}}{5}，$??$t_{2}=\frac {1-\sqrt {6}}{5}。$?

解：根據(jù)題意，得$(3m^{2}+4m - 3)+(-m^{2}+m - 30)=0，$

即$2m^{2}+5m - 33 = 0，$其中$a = 2，$$b = 5，$$c = -33，$$b^{2}-4ac=5^{2}-4×2×(-33)=25 + 264 = 289。$

則$m=\frac{-5\pm\sqrt{289}}{2×2}=\frac{-5\pm17}{4}，$

所以$m_{1}=\frac{-5 + 17}{4}=3，$$m_{2}=\frac{-5 - 17}{4}=-\frac{11}{2}。$

解：一元二次方程?$x^2-11x + 30 = 0，$?

其中?$a = 1，$??$b = -11，$??$c = 30，$??$b^2-4ac=(-11)^2-4×1×30 = 121 - 120 = 1。$?

則?$x=\frac {-(-11)\pm \sqrt {1}}{2×1}=\frac {11\pm 1}{2}，$?所以?$x_{1}=6，$??$x_{2}=5。$?

?$ $?當(dāng)?shù)妊切?$ABC$?的底邊長(zhǎng)為?$5、$?腰長(zhǎng)為?$6$?時(shí)，底邊上的高

?$h=\sqrt {6^2-(\frac {5}{2})^2}=\sqrt {36-\frac {25}{4}}=\sqrt {\frac {144 - 25}{4}}=\sqrt {\frac {119}{4}}=\frac {\sqrt {119}}{2}，$?

面積?$S=\frac {1}{2}×5×\frac {\sqrt {119}}{2}=\frac {5\sqrt {119}}{4}；$?

?$ $?當(dāng)?shù)妊切?$ABC$?的底邊長(zhǎng)為?$6、$?腰長(zhǎng)為?$5$?時(shí)，底邊上的高

?$h=\sqrt {5^2-(\frac {6}{2})^2}=\sqrt {25 - 9}=\sqrt {16}=4，$?

面積?$S=\frac {1}{2}×6×4 = 12。$?

綜上所述，?$\triangle ABC$?的面積為?$\frac {5\sqrt {119}}{4}$?或?$12。$?

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第13頁(yè)