`a)|2017x-2018|+|2017x+1|=2019` Áp dụng `|x|+|y|>=|x+y|,` ta được : `|2017x-2018|+|2017x+1|=||2018-2017x|+|2017x+1|>=|2018-2017x+2017x+1|=|2019|=2019` `->|2017x-2018|+|2017x+1|>=2019` Xảy ra dấu `=` khi : `(2017x-2018)(2017x+1)<=0` `->2017x-2018` và `2017x+1` trái dấu Mà `2017x+1>2017x-2018 AA x` `->{(2017x+1>=0),(2017x-2018<=0):}` `<=>{(2017x>=-1),(2017x<=2018):}` `<=>{(x>=(-1)/2017),(x<=2018/2017):}` Mà `x in Z -> {(x>=0),(x<=1):}->x in {0; 1}` Vậy : `x in {0; 1}` `b)A=1/3+1/4+1/5+...+1/130` `<=>A=(1/3+1/4)+(1/5+1/6+1/7+1/8)+(1/9+1/10+...+1/16)+(1/17+1/18+...+1/32)+(1/33+1/34+...+1/64)+(1/65+1/66+...+1/128)+(1/129+1/130)` `<=>A>(1/4+1/4)+(1/8+1/8+1/8+1/8)+(1/16+1/16+...+1/16)+(1/32+1/32+...+1/32)+(1/64+1/64+...+1/64)+(1/128+1/128+...+1/128)+(1/130+1/130)` `<=>A>1/(4)*2+1/(8)*4+1/(16)*8+1/(32)*16+1/(64)*32+1/(128)*64+1/(130)*2` `<=>A>1/2+1/2+1/2+1/2+1/2+1/2+1/115` `<=>A>1/2.6+1/115` `<=>A>3+1/115>3` `->A>3` Vậy : `A>3`
