Răspuns:
Explicație pas cu pas:
[tex]x+\dfrac{1}{|x+1|}=1|\cdot|x+1|\\x\cdot|x+1|+1=|x+1|\\x\cdot|x+1|-|x+1|+1=0\\|x+1|=\left \{ {{x+1,~x\geqslant -1} \atop {-x-1,~x<-1}} \right. \\\texttt{Cazul 1:}~x\geqslant -1\\x\cdot(x+1)-x-1+1=0\\x(x+1)-x=0\\x(x+1-1)=0\\x^2=0\Rightarrow x=0[/tex]
[tex]\texttt{Cazul 2:}~x<-1\\x(-x-1)-(-x-1)+1=0\\-x^2-x+x+1+1=0\\-x^2+2=0\\x^2=2\\|x|=\sqrt 2\\i)x=\sqrt 2~\texttt{nu convine, deoarece }\sqrt 2>-1\\ii)x=-\sqrt{2}~\texttt{solutia asta convine.}\\\texttt{Prin urmare, solutiile ecuatiei sunt:}\\\boxed{S=\{-\sqrt 2,0\}}[/tex]
