Reshite pojalusta))))))srocno❗️❗️❗️

$\sqrt{2x+8} =x \\\ \left \{ {{2x+8=x^2} \atop {x \geq 0}} \right. \\\ \left \{ {{x^2-2x-8=0} \atop {x \geq 0}} \right. \\\ \left \{ {{(x-4)(x+2)=0} \atop {x \geq 0}} \right. \\\ \left \{ {{x_1=4; \ x_2=-2} \atop {x \geq 0}} \right. \\\ x=4$
Ответ: 4

$\sqrt{5x+1} =1-x \\\ \left \{ {{5x+1=(1-x)^2} \atop {1-x \geq 0}} \right. \\\ \left \{ {{5x+1=1-2x+x^2} \atop {x \leq 1}} \right. \\\ \left \{ {{x^2-7x=0} \atop {x \leq 1}} \right. \\\ \left \{ {{x(x-7)=0} \atop {x \leq 1}} \right. \\\ \left \{ {{x_1=0; \ x_2=7} \atop {x \leq 1}} \right. \\\ x=0$
Ответ: 0