Question

уравнение, корень х-3=х-5

Answer 1

$\sqrt{x-3} =x-5\\ \left \{ {{x-3=(x-5) ^{2}/(1) } \atop {x-5 \geq 0}/(2)} \right. \\ (1)x-3=(x-5) ^{2} \\x-3= x^{2} +25-10x\\x-3- x^{2} -25+10x=0\\- x^{2} +11x-28=0/*(-1)\\ x^{2} -11x+28=0\\D=121-112=9\\ \sqrt{D} =3\\ x_{1} = \frac{11+3}{2} =7\\ x_{2} =4 \\ \\ (2)x-5 \geq 0\\x \geq 5$

Ответ:
$x=7$