用一元二次方程解:(1) 2x²-4x+1=0 (2) 2y²-7y+3=0 (3) 2x²-12x+7=0

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用一元二次方程解:(1) 2x²-4x+1=0 (2) 2y²-7y+3=0 (3) 2x²-12x+7=0

(1) 2x²-4x+1=0
x²-2x+1/2=0
x²-2x+1-1+1/2=0
(x-1)²=1/2
x-1=±√2/2
x=1±√2/2
(2)2y²-7y+3=0
(y-3)(2y-1)=0
y-3=0 或 2y-1=0,即 y=3 或 y=1/2
(3) 2x²-12x+7=0
x²-12x+36-36+7=0
(x-6)²=29
x-6=±√29
x=6±√29