一元二次方程的解法专题(自选)

1、专题训练（三）一元二次方程的解法1.用配方法解下列方程:2(1) (2x 1) = 1;(2) x + 4x 1 = 0;12(3) 4X 6x + 3 = 0.2 用公式法解下列方程:2(1) 5x + 2x 1 = 0;2(2) 6x + 13x + 6= 0;(3) x2+ 6x + 9= 7;2(4) 5x + 2 = 3x .3 .用因式分解法解下列方程:2(1) x 25= 0;2(2) x = 4x;(x 3)(x 1) = 6 2x.4.用适当的方法解下列方程：(1) (x 5)2= 16;2(2) x 3x = 5;2 2(3) (3x 4) = (4x 3);(2x 1)(

2、x + 1) = (3x + 1)(x + 1).5.解下列一元二次方程：2(1)x 4x 6= 0;2x 5x + 2= 0;(3)y(y 8) = 16 ;(4)4(x + 1)2= 9(x 2)2.1. (1)2x - 1= 1 ,Xi= 1, x2= 0.(2) 配方，得 x2+ 4x+ 4 = 5,即(x + 2)2= 5.开方，得 x+ 2 = 5.xi= 2 + 二.；5, X2= 2 J5.(3) x2- 24x + 12= 0，配方，得(x 12)2= 132. 开方，得 x 12= 2 33.x1= 2.j33+ 12 , x2= 2 33 + 12.22.(1)a=5,b

3、=2,c=1,b4ac=4+4X5x1=240,1 65.X2=c=6,b24ac=1694x6x6=250,32x2= 2，23-整理，得 x2+ 6x+ 2 = 0.Ta=1,b=6,c=2,b24ac=364x1x2=280,628丄厂 x= =37.X1= 3+ 茶 7, X2= 3 -J7.整理，得 3x2 5x 2= 0./a=3,b= 5,c= 2,b24ac=(5)24x3x(2)=490,549x=小6 -3.(1)(x + 5)(x 5) = 0, x+ 5 = 0，或 x 5= 0.X1= 5, X2= 5.(2) x2 4x = 0, x(x 4)= 0.x= 0,或

4、x 4 = 0.-X10, X24.(3) (x 3)(x 1) = 2(x 3), (x 3)(x 1 + 2) = 0.(x 3)(x + 1)= 0.x 3= 0，或 x+1 = 0.-X13, X21.4.(1)x5= 4,x=5 LX1=9,X2=1.(2)将原方程化为一般形式，得x2 3x 5= 0./a=1,b= 3,c= 5,b24ac=(3)24x1x(5)=9+20=2 90, (3) 29 329.x=2X12-(3)(3x 4)2 (4x 3)2= 0 , (3x 4+ 4x 3)(3x 4 4x+ 3) = 0.(7x 7)( x 1) =0. X1 , X21.(4)(x + 1)(2x 1 3x 1) = 0 , (x + 1)( x 2) = 0. x1 ,2.参考答案13 土 251213512(2)a= 6, b = 13,-X1=2,X2= 13.3+ .2925.(1)由原方程，得 x - 4x = 6.配方，得 X24x + 4 = 6+ 4，即(x 2)2= 10.直接开平方，得 X 2 = 10.解得 Xi= 2 + 10, X2= 2 10./ a=1,b=5,c=2,b24ac=(5)24x1X2=17,去括号，得 y8y =16 移项，得 y2 8y + 16= 0. 配方，得(y 4)2= 0.-y1=y2=4