Question

x+1 pe 2-x ≥0 rezolvati prin metoda intervalelor

Answer 1

Fractia este definita daca x≠2. Pentru x=-1 avem egalitate.
Acum ramane sa rezolvam inecuatia pentru cazul x≠-1.

In acest caz inegalitatea devine: $\frac{x+1}{2-x} \ \textgreater \ 0$ (egalitatea nu mai poate avea loc in acest caz)

$\frac{x+1}{2-x} \ \textgreater \ 0=\ \textgreater \ \left \{ {{x+1\ \textgreater \ 0} \atop {2-x\ \textgreater \ 0}} \right. ~sau~ \left \{ {{x+1\ \textless \ 0} \atop {2-x\ \textless \ 0}} \right. .$

$Cazul~I:~ \left \{ {{x+1\ \textgreater \ 0} \atop {2-x\ \textgreater \ 0}} \right. . \\ x+1\ \textgreater \ 0=\ \textgreater \ x\ \textgreater \ -1=\ \textgreater \ x\in(-1;+\infty). \\ 2-x\ \textgreater \ 0=\ \textgreater \ x\ \textless \ 2=\ \textgreater \ x\in(-\infty;;2). \\ Deci~x\in(-1;2). \\ \\ Cazul~II: \left \{ {{x+1\ \textless \ 0 \atop {2-x\ \textless \ 0}} \right. . \\ x+1\ \textless \ 0=\ \textgreater \ x\ \textless \ -1=\ \textgreater \ x\in(-\infty;-1). \\ 2-x\ \textless \ 0=\ \textgreater \ x\ \textgreater \ 2=\ \textgreater \ x\in(2;+\infty). \\ Acest~caz~nu~are~solutii.$

Solutia inecuatiei este (-1;2) ∪ {-1} =[-1;2).