Question

rezolvati si mie ce puteti din subiectul 1​

Answer 1

$\it \sqrt{x+4}-2= x \Rightarrow \sqrt{x+4}=x+2 \Rightarrow \begin{cases}\it x+4\geq0 \Rightarrow x\geq-4\\ \\ \it x+2\geq0 \Rightarrow x\geq-2\end{cases} \Rightarrow \\ \\ \\ \Rightarrow D=[-2,\ \infty)\\ \\ \sqrt{x+4}=x+2 \Rightarrow x+4=(x+2)^2 \Rightarrow \\ \\ \Rightarrow x+2+2=(x+2)^2 \Rightarrow (x+2)^2-(x+2)=2 \Rightarrow (x+2)(x+2-1)=2 \Rightarrow \\ \\ \Rightarrow (x+1)(x+2)=2 =1\cdot2=-2\cdot(-1) \Rightarrow \begin{cases}\it x+1=1 \Rightarrow x=0\in D\\ \\ x+1=-2 \Rightarrow x=-3<-2\end{cases}$

$\it Ecua\c{\it t}ia\ admite\ solu\c{\it t}ia\ unic\breve{a}\ x=0$