Question

数学题！2x^2-5x+1=0（用配方法）要详细的

Answer 1

2x^2-5x=-1
x^2-5x/2=-1/2
等号两边同加一次项系数一半的平方，一次项系数5/2得
x^2 -5x/2+（5/4）^2 = -1/2+(5/4）^2

( x-5/4 )^2 =17/16
x-5/4 =± (√17/4)

x1=（√17+5）/4
x2=（5 -√17）/4

Answer 2

2x^2-5x+1=0
2(x^2-5x/2+25/16)-25/8+1=0
2(x-5/4)^2=17/8
(x-5/4)^2=√17/16
x-5/4=±√17/4
x=±√17/4+5/4

Answer 3

温馨提示
熟悉配方法的原理这道题是不难的。原题是2x^2-5x+1=0
变形2（x^2-5x/2)+1=0，里面配方
2(x^2-5x/2+25/16)-25/16+1=0
2（x-5/4)^2=17/8
（x-5/4）^2=17/16
x-5/4=±√(17)/4
x=±√(17)/4+5/4

Answer 4

配方法
解：
x²-5/2x=-1/2
x²-5/2x+25/16=-1/2+25/16
(x-5/4)²=17/16
x-5/4=±√17/4
x=(5±√17)/4

Answer 5

2x^2-5x+1=2(x^2-5/2x)+1=2(x-5/4)^2-2*25/16+1=2(x-5/4)^2-17/8=0
(x-5/4)^2=17/16
x-5/4=±√17/4
x=(5±√17)/4