用适当方法解下列一元二次方程

2025-03-09 08:38:35
推荐回答(1个)
回答1:

2(x-3)²=x(x-3)
2(x-3)(x-3)-x(x-3)=0
(2x-6-x)(x-3)=0
(x-6)(x-3)=0
x1=6 x2=3

(x²-1)²-5(x²-1)+4=0
(x²-1-4)(x²-1-1)=0
x1=正负根号5 X2=正负根号2

(3x-4)²=(4x-3)²
(3x-4)²-(4x-3)²=0
(3x-4+4x-3)(3x-4-4x+3)=0
(7x-7)(-x-1)=0
7(x-1)(x+1)=0
x1=1 x2=-1