Question

stie cineva macar unul din ele?

Answer 1

3) $x \sqrt{x+1}-2 \sqrt{x+1}=2x-4 \\ \sqrt{x+1}(x-2)-2(x-2)=0 \\ (x-2)( \sqrt{x+1}-2)=0 \\ x-2=0 sau \sqrt{x+1}=2 \\ x=2saux+1=4 \\ x=2;x=3$

Answer 2

$3)\\ \\ x\sqrt{x+1}+4 =2\sqrt{x+1}+2x \\ \\ \\ \sqrt{x+1} = t \Rightarrow x+1 = t^2 \Rightarrow x = t^2-1,~~t \ \textgreater \ 0 \\ \\ \\ (t^2-1)\cdot t + 4 = 2t + 2(t^2-1) \\t(t^2-1)-2(t^2-1) -2t+4 = 0 \\ (t^2-1)(t-2) - 2(t-2) = 0 \\ (t-2)(t^2-1-2) = 0 \\ (t-2)(t^2-3) = 0 \\ \\ \bullet ~~ t = 2 \Rightarrow x = 2^2- 1 \Rightarrow x = 3 \\ \\ \bullet~~t^2-3 = 0 \Rightarrow t^2 = 3 \Rightarrow t = +\sqrt 3\Rightarrow x = \sqrt3^2-1 \Rightarrow \\ \\ \Rightarrow x = 2 \\ \\ \Rightarrow \boxed{S = \big\{2;3\big\}}$