Question

корень(16-x)+корень(x)=4

Answer 1

$\sqrt{16-x} +\sqrt{x}=4\\\\(\sqrt{16-x}+\sqrt{x})^{2} =4^{2} \\\\16-x+2\sqrt{x(16-x)}+x=16\\\\2\sqrt{x(16-x)} =0\\\\\sqrt{x(16-x)} =0 \\\\x(16-x)=0\\\\x_{1} =0\\\\16-x=0\\\\x_{2} =16$

Ответ : 0 ; 16