$\\$
Bài `3.`
`l,`
`(x+4)/2011 + (x+3)/2012 + (x+2)/2013 + (x+1)/2014 +4=0`
`-> ((x+4)/2011+1)+((x+3)/2012+1)+((x+2)/2013+1)+((x+1)/2014+1)=0`
`->(x+4+2011)/2011 +(x+3+2012)/2012 +(x+2+2013)/2013+(x+1+2014)/2014=0`
`->(x+2015)/2011+(x+2015)/2012+(x+2015)/2013+(x+2015)/2014=0`
`->(x+2015)(1/2011+1/2012+1/2013+1/2014)=0`
`->x+2015=0` (Vì `1/2011+1/2012+1/2013+1/2014 \ne 0)`
`->x=0-2015`
`->x=-2015`
Vậy `x=-2015`
$\\$
Bài `4.`
`c,`
`1/3-1/12-1/20-...-1/90-1/110 = x-5/13`
`->x-5/13=1/3 - (1/(3.4)+1/(4.5)+...+1/(9.10)+1/(10.11))`
`->x-5/13 = 1/3 - (1/3 - 1/4+1/4-1/5+...+1/9-1/10+1/10-1/11)`
`->x-5/13=1/3 - (1/3 - 1/11)`
`->x-5/13=1/3 - 1/3 +1/11`
`->x-5/13=1/11`
`->x=1/11+5/13`
`->x=68/143`
Vậy `x=68/143`
