Các câu hỏi tương tự
\(\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\)
\(\dfrac{x+1}{2004}+\dfrac{x+2}{2003}=\dfrac{x+3}{2002}+\dfrac{x+4}{2001}\)
1 tháng 3 2019 lúc 22:00
Giải phương trình:
\(\frac{2-x}{2002}-1=\frac{1-x}{2003}-\frac{x}{2004}\)
\(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
Giải phương trình sau:
a) x+1/2004 + x+2/2003 = x+3/2002 + x+4/2001
b) 201-x/99 + 203-x/97 + 205-x/95 + 3 = 0
Tính:
\(\frac{2-x}{2002}\)- 1 = \(\frac{1-x}{2003}-\frac{x}{2004}\)
giai phuong trinh
1.
(x+1)/2004+(x+2)/2003=(x+3)/2002+(x+4)/2001
2.(201-x)/99+(203-x)/97=(205-x)/95+3
a) \(\dfrac{x-45}{55}+\dfrac{x-47}{53}=\dfrac{x-55}{45}+\dfrac{x-53}{47}\)
b)\(\dfrac{x+1}{2004}+\dfrac{x+2}{2003}=\dfrac{x+3}{2002}+\dfrac{x+4}{2001}\)
\(\frac{X+1}{2004}+\frac{X+2}{2003}\) Bằng \(\frac{X+3}{2002}\:+\:\frac{x+4}{2001}\)