解一元二次方程：（1）（2x-3）2=(a-2)(3a-4) （2）（x2-2x-2)2-7(x2-2x-2)+6=0
2个回答
(1）（2x-3）2=(a-2)(3a-4)
4x-6=3a^2-10a+8
4x=3a^2-10a+14
x=(3a^2-10a+14)/4
（2）（x2-2x-2)2-7(x2-2x-2)+6=0
[(x^2-2x-2)-6][(x^2-2x-2)-1]=0
(x^2-2x-8)(x^2-2x-3)=0
(x-4)(x+2)(x-3)(x+1)=0
x=4,x=-2,x=3,x=-1
(3) x2+5x+k2=2kx+5k+6
x^2+x(5-2k)+k^2-5k-6=0
x^2+x(5-2k)+(k-6)(k+1)=0
[x-(k-6)][x-(k+1)]=0
x=k-6,x=k+1
(4)x2-4ax+3a2=1-2a
x^2-4ax+3a^2+2a-1=0
x^2-4ax+(3a-1)(a+1)=0
[x-(3a-1)][x-(a+1)]=0
x=3a-1,x=a+1
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