a) (x+1)/10+(x+1)/11+(x+1)/12=(x+1)/13+(x+1)/14
(x+1)/10+(x+1)/11+(x+1)/12-(x+1)/13-(x+1)/14=0
(x+1)(1/10+1/11+1/12-1/13-1/14)=0 (1)
thay 1/10>1/13
1/11>1/14
1/12>0
suy ra 1/10+1/11+1/12>1/13+1/14
suy ra 1/10+1/11+1/12-1/13-1/14>0
suy ra 1/10+1/11+1/12-1/13-1/14 khac 0
nền (1) tương dương x+1=0
tương dương x=-1
Vay x=-1
b) (x+4)/2000+(x+3)/2001=(x+2)/2002+(x+1)/2003
(x+4)/2000+(x+3)/2001-(x+2)/2002-(x+1)/2003=0
[(x+4)/2000+1]+[(x+3)/2001+1]-[(x+2)/2002+1]-[(x+1)/2003+1]=0
(x+2004)/2000+(x+2004)/2001-(x+2004)/2002-(x+2004)/2003=0
(x+2004)(1/2000+1/2001-1/2002/1/2003)=0 (2)
thay 1/2000>1/2002
1/2001>1/2003
suy ra 1/2000+1/2001>1/2002+1/2003
suy ra 1/2000+1/2001-1/2002-1/2003>0
suy ra 1/2000+1/2001-1/2002-1/2003 khac 0
nen (2) tuong duong x+2004=0
tuong duong x=-2004
Vay x=-2004
Trả lời :X = 2004.
Đúng thì k cho mk nha.