Trang chủ » Hỏi đáp » Môn Toán `(x-6)/1998 + (x-4)/2000 = (x-2000)/4 + (x-1998)/6` 01/01/2025 `(x-6)/1998 + (x-4)/2000 = (x-2000)/4 + (x-1998)/6`
(x-6)/1998+(x-4)/2000=(x-2000)/4+(x-1998)/6 =>(x-6)/1998+(x-4)/2000-2=(x-2000)/4+(x-1998)/6-2 =>((x-6)/1998-1)+((x-4)/2000-1)=((x-2000)/4-1)+((x-1998)/6-1) =>((x-6)/1998-1998/1998)+((x-4)/2000-2000/2000)=((x-2000)/4-4/4)+((x-1998)/6-6/6) =>(x-2004)/1998+(x-2004)/2000=(x-2004)/4+(x-2004)/6 =>(x-2004)/1998+(x-2004)/2000-(x-2004)/4-(x-2004)/6=0 =>(x-2004).(1/1998+1/2000-1/4-1/6)=0 Mà 1/1998+1/2000-1/4-1/6ne0 =>x-2004=0 =>x=2004 Vậy x=2004 Trả lời
(x-6)/1998+(x-4)/2000=(x-2000)/4+(x-1998)/6 <=>((x-6)/1998-1)+((x-4)/2000-1)=((x-2000)/4-1)+((x-1998)/6-1) <=>(x-2004)/1998+(x-2004)/2000=(x-2004)/4+(x-2004)/6 <=>(x-2004)(1/1998 +1/2000-1/4-1/6)=0 Do 1/1998 +1/2000-1/4-1/6\ne 0 =>x-2004=0 <=>x=2004 Vậy x=2004 Trả lời
2 bình luận về “`(x-6)/1998 + (x-4)/2000 = (x-2000)/4 + (x-1998)/6`”