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1. (1)$(x-3)^{2}-4=0$；
(2)$(2x-1)^{2}=25$；
(3)$3(x+2)^{2}=\dfrac{1}{3}$；
(4)$(3x-1)^{2}=(x+1)^{2}.$
(2)$(2x-1)^{2}=25$；
(3)$3(x+2)^{2}=\dfrac{1}{3}$；
(4)$(3x-1)^{2}=(x+1)^{2}.$
答案
(1)$x_1=1,x_2=5$
(2)$x_1=3,x_2=-2$
(3)$x_1=-\dfrac{5}{3},x_2=-\dfrac{7}{3}$
(4)$x_1=1,x_2=0$
(2)$x_1=3,x_2=-2$
(3)$x_1=-\dfrac{5}{3},x_2=-\dfrac{7}{3}$
(4)$x_1=1,x_2=0$
2. (1)$x^{2}-4x-1=0$;
(2)$x^{2}+3x-4=0$;
(3)$-\dfrac{1}{2}x^{2}+x+2=0$;
(4)$2x^{2}-5x+2=0$.
(2)$x^{2}+3x-4=0$;
(3)$-\dfrac{1}{2}x^{2}+x+2=0$;
(4)$2x^{2}-5x+2=0$.
答案
解:(1)移项,得$x^{2}-4x=1$,配方,得$x^{2}-4x+4=1+4$,即$(x-2)^{2}=5$,所以$x-2=\pm\sqrt{5}$,所以$x_1=2+\sqrt{5},x_2=2-\sqrt{5}$.
(2)移项,得$x^{2}+3x=4$,
配方,得$x^{2}+3x+(\dfrac{3}{2})^{2}=4+(\dfrac{3}{2})^{2}$,
即$(x+\dfrac{3}{2})^{2}=\dfrac{25}{4}$,
所以$x+\dfrac{3}{2}=\pm\dfrac{5}{2}$,所以$x_1=1,x_2=-4$.
(3)化二次项系数为1,得$x^{2}-2x-4=0$,
移项,得$x^{2}-2x=4$,
配方,得$x^{2}-2x+1=4+1$,
即$(x-1)^{2}=5$,所以$x-1=\pm\sqrt{5}$,
所以$x_1=1+\sqrt{5},x_2=1-\sqrt{5}$.
(4)化二次项系数为1,得$x^{2}-\dfrac{5}{2}x+1=0$,移项,得$x^{2}-\dfrac{5}{2}x=-1$,配方,得$x^{2}-\dfrac{5}{2}x+(\dfrac{5}{4})^{2}=-1+(\dfrac{5}{4})^{2}$,
即$(x-\dfrac{5}{4})^{2}=\dfrac{9}{16}$,所以$x-\dfrac{5}{4}=\pm\dfrac{3}{4}$,所以$x_1=2$,$x_2=\dfrac{1}{2}$.
(2)移项,得$x^{2}+3x=4$,
配方,得$x^{2}+3x+(\dfrac{3}{2})^{2}=4+(\dfrac{3}{2})^{2}$,
即$(x+\dfrac{3}{2})^{2}=\dfrac{25}{4}$,
所以$x+\dfrac{3}{2}=\pm\dfrac{5}{2}$,所以$x_1=1,x_2=-4$.
(3)化二次项系数为1,得$x^{2}-2x-4=0$,
移项,得$x^{2}-2x=4$,
配方,得$x^{2}-2x+1=4+1$,
即$(x-1)^{2}=5$,所以$x-1=\pm\sqrt{5}$,
所以$x_1=1+\sqrt{5},x_2=1-\sqrt{5}$.
(4)化二次项系数为1,得$x^{2}-\dfrac{5}{2}x+1=0$,移项,得$x^{2}-\dfrac{5}{2}x=-1$,配方,得$x^{2}-\dfrac{5}{2}x+(\dfrac{5}{4})^{2}=-1+(\dfrac{5}{4})^{2}$,
即$(x-\dfrac{5}{4})^{2}=\dfrac{9}{16}$,所以$x-\dfrac{5}{4}=\pm\dfrac{3}{4}$,所以$x_1=2$,$x_2=\dfrac{1}{2}$.
3. (1)$x^{2}-5x+1=0$；
(2)$x^{2}-2\sqrt{2}x+2=0$；
(3)$x(x+1)+4(x-1)=2(x-4)$；
(4)$x^{2}+mx-2m^{2}=0$($m$ 为常数).
(2)$x^{2}-2\sqrt{2}x+2=0$；
(3)$x(x+1)+4(x-1)=2(x-4)$；
(4)$x^{2}+mx-2m^{2}=0$($m$ 为常数).
答案
解:(1)$\because a=1,b=-5,c=1$,
$\therefore b^{2}-4ac=(-5)^{2}-4×1×1=21＞0$,
$\therefore x=\dfrac{-(-5)\pm\sqrt{21}}{2×1},\therefore x_1=\dfrac{5+\sqrt{21}}{2},x_2=\dfrac{5-\sqrt{21}}{2}$.
(2)$\because a=1,b=-2\sqrt{2},c=2$,
$\therefore b^{2}-4ac=(-2\sqrt{2})^{2}-4×1×2=0$,
$\therefore x=\dfrac{-(-2\sqrt{2})\pm0}{2×1}=\sqrt{2},\therefore x_1=x_2=\sqrt{2}$.
(3)化方程为一般形式,得$x^{2}+3x+4=0$.
$\because a=1,b=3,c=4$,
$\therefore b^{2}-4ac=3^{2}-4×1×4=9-16=-7＜0$,
$\therefore$此方程没有实数根.
(4)$\because a=1,b=m,c=-2m^{2}$,
$\therefore b^{2}-4ac=m^{2}-4×1×(-2m^{2})=9m^{2}$,
$\therefore x=\dfrac{-m\pm3m}{2×1},\therefore x_1=-2m,x_2=m$.
$\therefore b^{2}-4ac=(-5)^{2}-4×1×1=21＞0$,
$\therefore x=\dfrac{-(-5)\pm\sqrt{21}}{2×1},\therefore x_1=\dfrac{5+\sqrt{21}}{2},x_2=\dfrac{5-\sqrt{21}}{2}$.
(2)$\because a=1,b=-2\sqrt{2},c=2$,
$\therefore b^{2}-4ac=(-2\sqrt{2})^{2}-4×1×2=0$,
$\therefore x=\dfrac{-(-2\sqrt{2})\pm0}{2×1}=\sqrt{2},\therefore x_1=x_2=\sqrt{2}$.
(3)化方程为一般形式,得$x^{2}+3x+4=0$.
$\because a=1,b=3,c=4$,
$\therefore b^{2}-4ac=3^{2}-4×1×4=9-16=-7＜0$,
$\therefore$此方程没有实数根.
(4)$\because a=1,b=m,c=-2m^{2}$,
$\therefore b^{2}-4ac=m^{2}-4×1×(-2m^{2})=9m^{2}$,
$\therefore x=\dfrac{-m\pm3m}{2×1},\therefore x_1=-2m,x_2=m$.
4. (1)$5x^{2}-4x=0$；
(2)$x(x-6)=-4(x-6)$；
(3)$x^{2}-3x=x-3$；
(4)$4(2x+1)^{2}-9(2x-1)^{2}=0.$
(2)$x(x-6)=-4(x-6)$；
(3)$x^{2}-3x=x-3$；
(4)$4(2x+1)^{2}-9(2x-1)^{2}=0.$
答案
解:(1)原方程可化为$x(5x-4)=0$,所以$x=0$或$5x-4=0$,所以$x_1=0,x_2=\dfrac{4}{5}$.
(2)移项,得$x(x-6)+4(x-6)=0$,即$(x-6)(x+4)=0$,所以$x-6=0$或$x+4=0$,所以$x_1=6,x_2=-4$.
(3)移项,得$x(x-3)-(x-3)=0$,因式分解,得$(x-3)(x-1)=0$,则$x-3=0$或$x-1=0$,所以$x_1=3$,$x_2=1$.
(4)因式分解,得$[2(2x+1)+3(2x-1)][2(2x+1)-3(2x-1)]=0$,即$(4x+2+6x-3)(4x+2-6x+3)=0$,$(10x-1)(-2x+5)=0$,则$10x-1=0$或$-2x+5=0$,所以$x_1=\dfrac{1}{10},x_2=\dfrac{5}{2}$.
(2)移项,得$x(x-6)+4(x-6)=0$,即$(x-6)(x+4)=0$,所以$x-6=0$或$x+4=0$,所以$x_1=6,x_2=-4$.
(3)移项,得$x(x-3)-(x-3)=0$,因式分解,得$(x-3)(x-1)=0$,则$x-3=0$或$x-1=0$,所以$x_1=3$,$x_2=1$.
(4)因式分解,得$[2(2x+1)+3(2x-1)][2(2x+1)-3(2x-1)]=0$,即$(4x+2+6x-3)(4x+2-6x+3)=0$,$(10x-1)(-2x+5)=0$,则$10x-1=0$或$-2x+5=0$,所以$x_1=\dfrac{1}{10},x_2=\dfrac{5}{2}$.
