√(x-2+√(2x-5))+√(x+2+3√(2x-5))=7√2

$\sqrt{x-2+ \sqrt{2x-5} } + \sqrt{x+2+3 \sqrt{2x-5} } =7 \sqrt{2} \\ 3AMEHA:\ \sqrt{2x-5}=t,\ t \geq 0 \Rightarrow x=\dfrac{t^2+5}{2}\ \Rightarrow \\ \Rightarrow \sqrt{\dfrac{t^2+5}{2}-2+ t } + \sqrt{\dfrac{t^2+5}{2}+2+3 t} =7 \sqrt{2}\\ \sqrt{t^2+2t+1} + \sqrt{t^2+6t+9} =14\\ \sqrt{(t+1)^2} + \sqrt{(t+3)^2} =14\\ t+1+t+3=14\\ t=5 \Rightarrow x=\dfrac{5^2+5}{2}=15$

Проверка: $\sqrt{15-2+ \sqrt{30-5} } + \sqrt{15+2+3 \sqrt{30-5} } =7 \sqrt{2} \\ \sqrt{18} + \sqrt{32} =7 \sqrt{2} \\ 3 \sqrt{2} +4 \sqrt{2} =7 \sqrt{2} \\ 7 \sqrt{2} =7 \sqrt{2}$
Ответ: 15