(5+m)/(x-2)+1 = 1/(x-2) 无实数根,则m=多少?
问题描述：
(5+m)/(x-2)+1 = 1/(x-2) 无实数根,则m=多少?
答
(5+m)/(x-2)+1 = 1/(x-2) 移项：(5+m)/(x-2)+1 - 1/(x-2) = 0通分：[ (5+m) +（x-2) - 1 ] /(x-2) = 0(5+m +x-2- 1) /(x-2) = 0(x+2+m) /(x-2) = 0分母不为零,∴x≠2当x=2时无实数根2+2+m=0m=-4