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信息發(fā)布者:
證明:$A - B = 2a^2 - 4a - 1 - (a^2 - 2a - 4)$
$= 2a^2 - 4a - 1 - a^2 + 2a + 4$
$= a^2 - 2a + 3$
$= (a - 1)^2 + 2$
$\because (a - 1)^2 \geq 0$
$\therefore (a - 1)^2 + 2 > 0,$即$A - B$的值恒為正數(shù)。
解:$2x^2 - 7x - 4 = 0$
$x^2 - \frac{7}{2}x = 2$
$x^2 - \frac{7}{2}x + \frac{49}{16} = 2 + \frac{49}{16}$
$(x - \frac{7}{4})^2 = \frac{81}{16}$
$x - \frac{7}{4} = \pm \frac{9}{4}$
$x_1 = 4,$$x_2 = -\frac{1}{2}$
解:$3y^2 - 6y - 1 = 0$
$y^2 - 2y = \frac{1}{3}$
$y^2 - 2y + 1 = \frac{1}{3} + 1$
$(y - 1)^2 = \frac{4}{3}$
$y - 1 = \pm \frac{2\sqrt{3}}{3}$
$y_1 = 1 + \frac{2\sqrt{3}}{3},$$y_2 = 1 - \frac{2\sqrt{3}}{3}$
解:$-\frac{1}{2}x^2 - \frac{1}{2}x + 2 = 0$
兩邊同乘$-2$得:$x^2 + x - 4 = 0$
$x^2 + x = 4$
$x^2 + x + \frac{1}{4} = 4 + \frac{1}{4}$
$(x + \frac{1}{2})^2 = \frac{17}{4}$
$x + \frac{1}{2} = \pm \frac{\sqrt{17}}{2}$
$x_1 = -\frac{1}{2} + \frac{\sqrt{17}}{2},$$x_2 = -\frac{1}{2} - \frac{\sqrt{17}}{2}$
解:$\sqrt{2}x^2 - 4x + 4\sqrt{2} = 0$
兩邊同除以$\sqrt{2}$得:$x^2 - 2\sqrt{2}x + 4 = 0$
$x^2 - 2\sqrt{2}x = -4$
$x^2 - 2\sqrt{2}x + 2 = -4 + 2$
$(x - \sqrt{2})^2 = -2$
$\because (x - \sqrt{2})^2 \geq 0$
$\therefore$原方程無解
A
【答案】:
解:$x^2-\frac 72x=2$
$ x^2-\frac 72x+\frac {49}{16}=\frac {81}{16}$
$ \ \ \ \ (x-\frac 74)^2=\frac {81}{16}$
$ \ \ \ \ \ \ \ x-\frac 74=±\frac 94$
$ x_1=4,$$x_2=-\frac 12$
解:$y^2-2y-\frac 13=0$
$ \ \ \ \ \ y^2-2y+1=\frac 43$
$ \ \ \ \ \ \ \ \ (y-1)^2=\frac 43$
$ \ \ \ \ \ \ \ \ \ \ y-1=±\frac {2\sqrt {3}}3$
$ y_1=1+\frac {2\sqrt {3}}3,$$y_2=1-\frac {2\sqrt {3}}3$
解:$x^2+x-4=0$
$        x^2+x+\frac 14=\frac {17}4$
$            (x+\frac 12)^2=\frac {17}4$
$                x+\frac 12=±\frac {\sqrt {17}}2$
$ x_1=-\frac 12+\frac {\sqrt {17}}2,$$x_2=-\frac 12-\frac {\sqrt {17}}2$
解:$x^2-2\sqrt {2}x+4=0$
$ x^2-2\sqrt {2}x+2=-2$
$ (x-\sqrt {2})^2=-2$
∵$(x-\sqrt {2})^2≥0$
∴原方程無解

【解析】:
(1)解:$2x^{2}-7x-4=0$,$a=2$,$b=-7$,$c=-4$,$\Delta=(-7)^{2}-4×2×(-4)=49+32=81$,$x=\frac{7\pm\sqrt{81}}{2×2}=\frac{7\pm9}{4}$,$x_{1}=4$,$x_{2}=-\frac{1}{2}$;
(2)解:$a=3$,$b=-6$,$c=-1$,$\Delta=(-6)^{2}-4×3×(-1)=36+12=48$,$y=\frac{6\pm\sqrt{48}}{2×3}=\frac{6\pm4\sqrt{3}}{6}=1\pm\frac{2\sqrt{3}}{3}$,$y_{1}=1+\frac{2\sqrt{3}}{3}$,$y_{2}=1-\frac{2\sqrt{3}}{3}$;
(3)解:方程兩邊同乘$-2$得$x^{2}+x-4=0$,$a=1$,$b=1$,$c=-4$,$\Delta=1^{2}-4×1×(-4)=1+16=17$,$x=\frac{-1\pm\sqrt{17}}{2×1}$,$x_{1}=\frac{-1+\sqrt{17}}{2}$,$x_{2}=\frac{-1-\sqrt{17}}{2}$;
(4)解:$a=\sqrt{2}$,$b=-4$,$c=4\sqrt{2}$,$\Delta=(-4)^{2}-4×\sqrt{2}×4\sqrt{2}=16-32=-16<0$,原方程沒有實數(shù)根.
【答案】:
A

【解析】:
解:$3x^{2}-6x+1=0$
$3x^{2}-6x=-1$
$x^{2}-2x=-\frac{1}{3}$
$x^{2}-2x+1=-\frac{1}{3}+1$
$(x-1)^{2}=\frac{2}{3}$
結(jié)論:A