Posted Sat Nov 14, 2015 10:53 pm
$ <=>\sqrt { x } (\sqrt { x+1 } +\sqrt { x+2 } -\sqrt { x+3 } )=0\\ <=>x=0\quad hoặc\quad \sqrt { x+1 } +\sqrt { x+2 } -\sqrt { x+3 } =0\\ =>2x+3+2\sqrt { (x+1)(x+2) } =x+3\\ <=>2\sqrt { (x+1)(x+2) } =-x(x\le 0)\\ <=>4({ x }^{ 2 }+3x+2)={ x }^{ 2 }<=>3{ x }^{ 2 }+12x+8=0\\ <=>x=\dfrac { -6\pm 2\sqrt { 3 } }{ 3 } $