Question

Va rog ajutati-ma!Rezolvati inecuatiile:$\frac{x+1}{x+2} \leq 2$

Answer 1

$\displaystyle \mathtt{ \frac{x+1}{x+2} \leq 2 }\\ \\ \mathtt{ \frac{x+1}{x+2}-2 \leq 0 }\\ \\ \mathtt{ \frac{x+1-2(x+2)}{x+2} \leq 0}\\ \\ \mathtt{ \frac{x+1-2x-4}{x+2} \leq 0}\\ \\ \mathtt{ \frac{-x-3}{x+2} \leq 0 }\\ \\ \mathtt{ \frac{x+3}{x+2} \geq 0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\mathtt{x+3=0 \Rightarrow x=-3}}~\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x+2=0 \Rightarrow x=-2}$

$\displaystyle \mathtt{x~~~~~~~~~~~~~~~-\infty~~~~~~~~~~~~~~~-3~~~~~~~~~~~~~~~~~-2~~~~~~~~~~~~~~~~~~~~~+\infty}\\ \\ \mathtt{x+3~~~~~~~~~--------~~0++++++++++++++++}\\ \\ \mathtt{x+2~~~~~~~~~~----~-----------~0+++++++++}\\ \\ \mathtt{ \frac{x+3}{x+2} ~~~~~~~~~++++++++~0-------|+++++++++} \\ \\ \\ \mathtt{x \in (-\infty,-3]\cup(-2,+\infty)}$