2 câu trả lời
1/3+1/6+...+2/(x(x+1))=2013/2015(x\ne 0,x\ne -1)
->2/6 +2/12+...+2/(x(x+1))=2013/2015
-> 2/(2.3) +2/(3.4)+...+2/(x(x+1))=2013/2015
-> 2 (1/2-1/3+1/3-1/4+...+1/x-1/(x+1))=2013/2015
->1/2 - 1/(x+1)=2013/4030
->1/(x+1)=1/2015
->x+1=2015
->x=2014 (Tm)
Vậy x=2014
1/3+1/6+1/10+...+2/(x(x+1))=2013/2015(x\ne0; x\ne-1)
<=>2/6+2/12+2/20+...+2/(x(x+1))=2013/2015
<=>2[1/6+1/12+1/20+...+1/(x(x+1))]=2013/2015
<=>2[1/2.3+1/3.4+1/4.5+...+1/(x(x+1))]=2013/2015
<=>2(1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/(x+1))=2013/2015
<=>2(1/2-1/(x+1))=2013/2015
<=>1-2/(x+1)=2013/2015
<=>2/(x+1)=2/2015
=>x+1=2015
<=>x=2014(TM)
Vậy x=2014
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