Question

√2x-4/x+5 (tot sub radical) ∈Ζ 
aflati numerele intregi x

VA ROG REZOLVARE COMPLETA

Answer 1

$\sqrt{ \frac{2x-4}{x+5}} = \sqrt{ \frac{2x+10-10-4}{x+5}} = \\ \\ =\sqrt{ \frac{2x+10-14}{x+5}} =\sqrt{ \frac{2x+10}{x+5}+\frac{-14}{x+5}} = \\ \\ =\sqrt{ \frac{2(x+5)}{x+5}+\frac{-14}{x+5}} =\sqrt{ 2+\frac{-14}{x+5}} \\ \\ D_{-14}=\{ -14; -7;-2;-1;1;2;7;14\} \\ \text{Impartind pe -14 la fiecare din divizori obtinem multimea caturilor:} \\ \{1;2;7;14;-14;-7;-2;-1\} \\ \text{Dintre acestea, urmatoarele caturi dau patrat perfect sub radical:} \\ \{ 2;14;-2;-1\} \\ => \text{avem 4 solutii}$


$Solutia\;1: \\ \frac{-14}{x+5}=2 =>x+5= -7 =>x=-5-7=\boxed{-12} \\ =>\sqrt{ 2+\frac{-14}{x+5}}=\sqrt{ 2+\frac{-14}{-12+5}}=\sqrt{ 2+\frac{-14}{-7}}=\sqrt{ 2+2}=\sqrt{ 4}=2 \\ \\ Solutia\;2: \\ \frac{-14}{x+5}=14 =>x+5= -1 =>x=-1-5=\boxed{-6} \\ =>\sqrt{ 2+\frac{-14}{x+5}}=\sqrt{ 2+\frac{-14}{-6+5}}=\sqrt{ 2+\frac{-14}{-1}}=\sqrt{ 2+14}=\sqrt{ 16}=4$


$Solutia\;3: \\ \frac{-14}{x+5}=-2 =>x+5= 7 =>x=7-5=\boxed{2} \\ =>\sqrt{ 2+\frac{-14}{x+5}}=\sqrt{ 2+\frac{-14}{2+5}}=\sqrt{ 2+\frac{-14}{7}}=\sqrt{ 2-2}=\sqrt{ 0}=0 \\ \\ Solutia\;4: \\ \frac{-14}{x+5}=-1 =>x+5= 14 =>x=14-5=\boxed{9} \\ =>\sqrt{ 2+\frac{-14}{x+5}}=\sqrt{ 2+\frac{-14}{9+5}}=\sqrt{ 2+\frac{-14}{14}}=\sqrt{ 2-1}=\sqrt{ 1}=1$