Question

Уравнение решить корень из (х-9)-корень из(х-16)=1

Answer 1

ОДЗ
$\left \{ {{x-9 \geq 0} \atop {x-16 \geq 0}} \right. <= \left \{ {{x \geq 9} \atop {x \geq 16}} \right. \\\ [16; +\infty)$
$\sqrt{x-9}-\sqrt{x-16}=1$
$(\sqrt{x-9}-\sqrt{x-16})^2=1^2\\\ (\sqrt{x-9})^2-2\sqrt{x-9}\sqrt{x-16}+(\sqrt{x-16})^2=1\\\ x-9-2\sqrt{x^2-25x+144}+x-16=1\\\ -2\sqrt{x^2-25x+144}=26-2x\\\ \sqrt{x^2-25x+144}=x-13\\\ (\sqrt{x^2-25x+144})^2=(x-13)^2\\\ x^2-25x+144=x^2-26x+169\\\ x=25$