解二元一次方程2x^2-8x+4=0
问题描述:
解二元一次方程2x^2-8x+4=0
答
=x^2-4x+2=x^2-4x+4-2=====>(x-2)^2=2 x=2+正负根2
答
x=正根号2或负根号2
答
2x^2-8x+4=0
x^2-4x+2=0
x^2-4x+4-2=0
x^2-4x+4=2
(x-2)²=2
x-2=±√2
x=2±√2