Решите пожалуйста уравнение:

Решите задачу:

$\sqrt{x+6}+\sqrt{x-1}+2*\sqrt{x^2+5x-6}= 51-2x\\\\ \sqrt{x+6}+\sqrt{x-1}+2*\sqrt{(x+6)(x-1)}= 51+6-1-x-6-x+1\\\\ \sqrt{x+6}+\sqrt{x-1}+2*\sqrt{x+6}*\sqrt{x-1}= 56-(x+6)-(x-1)\\\\$

$0 \leq T=\sqrt{x+6}\ \ \ \ 0 \leq U=\sqrt{x-1}\\\\ T+U+2*T*U= 56-T^2-U^2\\\\ T^2+2*T*U+U^2+T+U=56\\\\ (T+U)^2+T+U=56\\\\ 0\ \textless \ W=T+U\\\\ W^2+W-56=0\\\\ W=-8\ \ or\ \ W=+7\\\\ ----------------------------\\ T+U=7\\\\\\ \left \{ {{T+U=7} \atop {T^2-U^2=7}} \right. ; \left \{ {{T+U=7} \atop {(T+U)(T-U)=7}} \right.; \left \{ {{T+U=7} \atop {T-U=1}} \right. \\\\$

$\sqrt{x+6}-\sqrt{x-1}=1\\\\ \sqrt{x+6}=1+\sqrt{x-1}\\\\ \left \{ {{x+6=(1+\sqrt{x-1})^2} \atop {1+\sqrt{x-1} \geq 0}} \right. \\\\ x+6=1+2\sqrt{x-1}+x-1}\\\\ 3=\sqrt{x-1}\\\\ \left \{ {{x-1=3^2} \atop {3 \geq 0}} \right. \\\\ x=3^2+1\\\\ x=10$