[tex]\it Not\breve am:\ \ \sqrt x+\sqrt{x+1}=a,\ \ \sqrt x+\sqrt{x+2}=b,\ \ evident\ a,\ b > 0, \ a < b\\ \\ 2log_4 b=2\cdot\dfrac{log_2 b}{log_2 4}=2\cdot\dfrac{log_2 b}{2}=log_2 b\\ \\ \\ Ecua\c{\it t}ia\ devine:\ \ log_2 a+log_2 b=\dfrac{1}{2} \Rightarrow log_2a\cdot b=\dfrac{1}{2} \Rightarrow a\cdot b= 2^{\frac{1}{2}} \Rightarrow[/tex]
[tex]\it \Rightarrow a\cdot b=\sqrt2 =1\cdot\sqrt2 \Rightarrow \begin{cases} \it a=1 \Rightarrow \sqrt x+\sqrt{x+1}=1 \Rightarrow x=0\\ \\ \it b=\sqrt2 \Rightarrow \sqrt x+\sqrt{x+2}=\sqrt2\Rightarrow x=0\end{cases}\\ \\ x=0\ verific\breve a\ \d si\ ecua\c{\it t}ia\ ini\c{\it t}ial\breve a.\\ \\ Prin\ urmare,\ ecua\c{\it t}ia\ dat\breve a\ admite\ solu\c{\it t}ia\ x=0.[/tex]
