Ta có: `|5-x|=5-x` khi `5-xge0` `=>xle5` `|5-x|=-5+x` khi `5-x<0=>x>5` Trường hợp `1`:`xle5` `|5-x|=3x+1` `=>5-x=3x+1` `=>-x-3x=1-5` `=>-4x=-4` `=>x=-4:(-4)` `=>x=1(TMĐK:xle5)` Trường hợp `2`:`x>5` `|5-x|=3x+1` `=>5-x=-(3x+1)` `=>5-x=-3x-1` `=>-x+3x=-1-5` `=>x.(-1+3)=-6` `=>x.2=-6` `=>x=-6:2` `=>x=-3(KTMĐK:x>5)` Vậy `x=1`
