(3/x)+1/(x-1)+1/(x-2)+1/(x-3)+1/(x-4)+3/(x-5)=0 [(3/x)+3/(x-5)]+[1/(x-1))+1/(x-4)]+[1/(x-2)+3/(x-3)]=0 [3(2x-5)/x(x-5)]+[(2x-5)/(x-1)(x-4)]+[4(2x-5)/(x-2)(x-3)]=0 (2x-5)[3/x(x-5)+1/(x-1)(x-4)+4/(x-2)(x-3)]=0 2x-5=0 ===>x1=5/2 或3/x(x-5)+1/(x-1)(x-4)+1/(x-2)(x-3) =0 ==> 3/(x²-5x) +1/(x²-5x+4) +4/(x²-5x+6) =0 换元,x²-5x+4=T ==>3/(T-3) +1/T +4/(T+2) =0 ==>2T²-3T-2=0 ==>T=2,T=-1/2 T=2 ===>x =(5 ±√17)/2 T=-1/2 ==>x=(5 ±√7)/2 显然上述5个解都不是增根 ==>方程的解 x1=5/2 ; x2,3 =(5 ±√17)/2 x4,5=(5 ±√7)/2 。
(3/x)+1/(x-1)+1/(x-2)+1/(x-3)+1/(x-4)+3/(x-5)=0 [(3/x)+3/(x-5)]+[1/(x-1))+1/(x-4)]+[1/(x-2)+1/(x-3)]=0 [3(2x-5)/x(x-5)]+[(2x-5)/(x-1)(x-4)]+[(2x-5)/(x-2)(x-3)]=0 (2x-5)[3/x(x-5)+1/(x-1)(x-4)+1/(x-2)(x-3)]=0 2x-5=0 x=5/2
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答：(1)X-3>0，X+1<0或X-3<0，X+1>0即可 答案是（-1,3） （2）要分X的取值，分别有三个，①X=-1,P为空集符合条件 ②X>-1，P为（-...详情>>
答：我会！！！ 选D 用选择题嘛 用排除法就可以做出来的详情>>
