Question

ПОМОГИТЕ С 1 УРОВНЕНИЕМ СРОЧНО , ДАЮ БАЛЛЫ!!!

Answer 1

$\displaystyle\bf\\ODZ \ :\\\\\left \{ {{2x+1\geq 0} \atop {x\geq 0}} \right. \ \ \ \Rightarrow \ \ \ \left \{ {{x\geq -0,5} \atop {x\geq 0}} \right. \ \ \ \Rightarrow \ \ \ \boxed{x\geq 0}\\\\\\\sqrt{2x+1} -\sqrt{x} =1\\\\\\\sqrt{2x+1} =1+\sqrt{x} \\\\\\\Big(\sqrt{2x+1} \Big)^{2} =\Big(1+\sqrt{x} \Big)^{2} \\\\\\2x+1=1+2\sqrt{x} +x\\\\\\2x+1-1-2\sqrt{x} -x=0\\\\\\x-2\sqrt{x} =0\\\\\\(\sqrt{x} )^{2} -2\sqrt{x} =0\\\\\\\sqrt{x} \cdot(\sqrt{x} -2)=0$

$\displaystyle\bf\\\left[\begin{array}{ccc}\sqrt{x} =0\\\sqrt{x} -2=0\end{array}\right\\\\\\\left[\begin{array}{ccc}x=0\\\sqrt{x} =2\end{array}\right\\\\\\\left[\begin{array}{ccc}x_{1} =0\\x_{2} =4\end{array}\right\\\\\\Otvet \ : \ 0 \ ; \ 4$