3)
b)
Consider ca la punctul b) este o greseala de tipar sau de editare a cartii.
x nu se inmulteste cu paranteza.
Tinand cont de fractiile din paranteza si de fractiile de dupa egal,
consider ca x se aduna cu paranteza.
Rezolvare:
[tex]\displaystyle\\x+\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right)=2019-\frac{1}{2}-\frac{2}{3}-\frac{3}{4}-...-\frac{2018}{2019}\\\\\\x+\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right)-\left(-\frac{1}{2}-\frac{2}{3}-\frac{3}{4}-...-\frac{2018}{2019}\right)=2019\\\\\\x+\left(\underbrace{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}}_{2019~termeni}\right)+\left(+\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{2018}{2019}\right)=2019[/tex]
.
Desfacem parantezele si imperechem fractiile doua cate doua.
.
[tex]\displaystyle\\x+1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}+\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{2018}{2019}=2019\\\\\\x+1+\Big(\frac{1}{2}+\frac{1}{2}\Big)+\Big(\frac{1}{3}+\frac{2}{3}\Big)+\Big(\frac{1}{4}+\frac{3}{4}\Big)+...+\Big(\frac{1}{2019}+\frac{2018}{2019}\Big)=2019\\\\\\x+\underbrace{1+\frac{2}{2}+\frac{3}{3}+\frac{4}{4}+...+\frac{2019}{2019}}_{\bf 2019~termeni}=2019\\\\\\x+1+1+1+1+...+1=2019\\\\x+2019=2019\\\\\boxed{\bf x=0}[/tex]
