1. điều kiện:
30x≥0⇒x≥0
ta có:
|x+1/3|+|x+1/15|+|x+1/35|+...+|x+1/2499|=30x
=> x+1/3+x+1/15+...+x+1/2499=30x
=>25x+(1/3+1/15+...+1/2499)=30x
=>1/(1.3)+1/(3.5)+...+1/(49/50)=5x
=>2/(1.3)+2/(3.5)+...+2/(49.50)=10x
=>1-(1/3)+1/3-1/5+...+1/49-1/50=10x
=> 1-1/50=10x
=>49/50=10x
=> x=49/500
2. TA có:
A=|x-1+|x-2|+|x-3|+|x-4|+2016
=(|x-1|+|x-4|)+(|x-2|+|x-3|)+2016
=(|x-1|+|4-x|)+(|x-2|+|3-x|)+2016
|x-1+4-x|+|x-2+3-x|+2016
=3+1+2016
=2020
=> Min A=2020⇔$\left \{ {{(x-1)(4-x)≥0} \atop {(x-2)(3-x)≥0}} \right.$ =>$\left \{ {{4≥x≥1} \atop {3≥x≥2}} \right.$
vậy Min A=2020⇔3≥x≥2
