Question

Решить уравнение x+√x=0

Answer 1

Ответ:

0.

Объяснение:

$x+\sqrt{x} =0;\\\sqrt{x} =-x;\\\left \{ \begin{array}{lcl} {({\sqrt{x})^{2}  =(-x)^{2}, } \\ {-x\geq 0;}} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{x=x^{2} ,} \\ {x\leq 0;}} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{x-x^{2} =0,} \\ {x\leq 0;}} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{x(1-x)=0,} \\ {x\leq0; }} \end{array} \right.\Leftrightarrow$$\left \{ \begin{array}{lcl} {{\left [ \begin{array}{lcl} {{x=0,} \\ {x=1}} \end{array} \right.} \\ {x\leq 0;}} \end{array} \right.\Leftrightarrow x =0.$