Question

$Problema~: \\ Multimea~solutiilor~inecuatiei: \\ (1+ \frac{x-1}{x+3} )(2+ \frac{x-2}{x+1} )\ \textgreater \ 0 \\ Care~e???$

Answer 1

$Domeniul~de~de.fi.ni.ti.e: ~ x \not \in \{-3;-1 \}. \\ \\ \Big(1+ \frac{x-1}{x+3 } \Big) \Big( 2+ \frac{x-2}{x+1} \Big)\ \textgreater \ 0 \\ \\ \frac{2x+2}{x+3} \cdot \frac{3x}{x+1}\ \textgreater \ 0 \\ \\ \frac{2(x+1)}{x+3} \cdot \frac{3x}{x+1}\ \textgreater \ 0 \\ \\ \frac{6x}{x+3}\ \textgreater \ 0.~Si~impartind~la~6~(6\ \textgreater \ 0,~important!) \Rightarrow \frac{x}{x+3}\ \textgreater \ 0. \\ \\ Deci~x~si~x+3~au~acelasi~semn. \\ \\ Cazul~1:~x\ \textless \ 0,~atunci~si~x+3\ \textless \ 0,~deci~x\ \textless \ -3 \\ \\ Cazul~2:~x\ \textgreater \ 0.~In~acest~caz~x+3\ \textgreater \ 3\ \textgreater \ 0,~deci~avem~simultan \\ \\ x\ \textgreater \ 0~si~x+3\ \textgreater \ 0 .$

$Deci~x\ \textgreater \ 0~este~solutie~pentru~cazul~2. \\ \\ SOLUTIE:~(- \infty;-3) \cup (0;+ \infty)~(adica~R-[-3;0]).$