x+x/2+(x+1)/3+(x+2)/4+…..+(x+n)/(n+2)=n+3
x+x/2+x/3+x/4+….+x/(n+2)+1/3+2/4+3/5+…+n/(n+2)=n+3
x(1+1/2+1/3+1/4+…..+1/(n+2)=n+3-[1/3+2/4+3/5+…+n/(n+2)]
În partea dreapta sunt n fracții, am repartizat cate un 1, (din cei n), la fiecare fracție.
x(1+1/2+1/3+1/4+…..+1/(n+2)=3+(1 -1/3)+(1 -2/4)+(1 -3/5)+…+[1 - n/(n+2)]
x(1+1/2+1/3+1/4+…..+1/(n+2)=3+2/3+2/4+…..+2/(n+2)
x(1+1/2+1/3+1/4+…..+1/(n+2)=2[3/2+1/3+1/4+…..+1/(n+2)]
x(1+1/2+1/3+1/4+…..+1/(n+2)=2[1+1/2+1/3+1/4+…..+1/(n+2)]
=> x=2
