Question

\frac{x-1}{x+2} > \frac{x-2}{x+1} +3

Answer 1

$\frac{x-1}{x+2} \ \textgreater \ \frac{x-2}{x+1} +3 \ \textless \ =\ \textgreater \ \frac{x-1}{x+2} \ \textgreater \ \frac{x-2+3x+3}{x+1} \ \textless \ =\ \textgreater \ \frac{x-1}{x+2} \ \textgreater \ \frac{4x+1}{x+1}$
inmultim cu (x+2)(x+1)
(x-1)(x+1) > (4x+1)(x+2) <=> $x^{2} -1 \ \textgreater \ 4 x^{2} +8x+x+2 \ \textless \ =\ \textgreater \ -3x^{2} -9x-3 \ \textgreater \ 0 \ \textless \ =\ \textgreater \ 3x^{2} +9x+3$$\ \textless \ 0 \ \textless \ =\ \textgreater$ 
Δ=45
$\left \{ {{x_{1} =- \frac{3+ \sqrt{5} }{2} } \atop {x_{2} =- \frac{3- \sqrt{5} }{2}} \right.$
Facem graficul si observam pe care intervale x<0
R-s: x ∈ {$- \frac{3+ \sqrt{5} }{2}; - \frac{3- \sqrt{5} }{2}$}