求解一道行列式,谢谢记行列式x-2x-1x-2x-32x-22x-12x-22x-3为f(x),则方程f(x)=0的根的个数为.3x-33x-24x-53x-5答案为：3求高手详细解答,

Question

求解一道行列式,谢谢
记行列式
x-2 x-1 x-2 x-3
2x-2 2x-1 2x-2 2x-3 为f(x),则方程f(x)=0的根的个数为__________.
3x-3 3x-2 4x-5 3x-5 答案为：3 求高手详细解答,谢谢!
4x 4x-3 5x-7 4x-3

Answer 1

x-2,x-1,x-2,x-3
2x-2,2x-1,2x-2,2x-3
3x-3,3x-2,4x-5,3x-5
4x,4x-3,5x-7,4x-3
r2-[(2x-2)/(x-2)]r2:
x-2,x-1,x-2,x-3
0,-x/(x-2),0,x/(x-2)
3x-3,3x-2,4x-5,3x-5
4x,4x-3,5x-7,4x-3
r3-[(3x-3)(x-2)]r1:
x-2,x-1,x-2,x-3
0,-x/(x-2),0,x/(x-2)
0,-(2x-1)/(x-2),x-2,(x+1)/(x-2)
4x,4x-3,5x-7,4x-3
r4-[4x/(x-2)]r1:
x-2,x-1,x-2,x-3
0,-x/(x-2),0,x/(x-2)
0,-(2x-1)/(x-2),x-2,(x+1)/(x-2)
0,-(7x-6)/(x-2),x-7,(x+6)/(x-2)
r3-[(2x-1)/x]r2:
x-2,x-1,x-2,x-3
0,-x/(x-2),0,x/(x-2)
0,0,x-2,-1
0,-(7x-6)/(x-2),x-7,(x+6)/(x-2)
r4-[(7x-6)/x]r2:
x-2,x-1,x-2,x-3
0,-x/(x-2),0,x/(x-2)
0,0,x-2,-1
0,0,x-7,-6
r4-(x-7)/(x-2)r3:
x-2,x-1,x-2,x-3
0,-x/(x-2),0,x/(x-2)
0,0,x-2,-1
0,0,0,-5(x-1)/(x-2)
=(x-2)*[-x/(x-2)]*(x-2)*[-5(x-1)/(x-2)]
=(-x)*[-5(x-1)]
=5x^2-5x
依题意得：f(x)=0
即：5x^2-5x=0
x=0或x=1
已经过计算机验证,计算没有错,我得出的答案是2,不知道答案的3是怎么来的.