Question

Розвязати рівняння дуже прошу!!!v

Answer 1

Ответ:

$1)\ \ \sqrt{x-2}-3\sqrt[4]{x-2}-4=0\ \ ,\ \ \ \ ODZ:\ x\geq 2\ ,\\\\t=\sqrt[4]{x-2}\geq 0\ \ ,\ \ \ t^2-3t-4=0\ \ ,\ \ t_1=-1\ ,\ t_2=4\ \ (teorema\ Vieta)\\\\t=\sqrt[4]{x-2}=4\geq 0\ \ ,\ \ x-2=4^4\ \ ,\ \ x-2=256\ \ ,\ \ \boxed{\ x=258\ }$

$2)\ \ \sqrt{x+6}-\sqrt{7-x}=1\ \ ,\ \ \ ODZ:\ \left\{\begin{array}{l}x+6\geq 0\\7-x\geq 0\end{array}\right\ \ \left\{\begin{array}{l}x\geq -6\\x\leq 7\end{array}\right\\\\-6\leq x\leq 7\\\\\sqrt{x+6}=1+\sqrt{7-x}\\\\x+6=1+2\sqrt{7-x}+7-x\\\\2\sqrt{7-x}=x+6+x-8\\\\\sqrt{7-x}=x-1\\\\7-x=x^2-2x+1\\\\x^2-x-6=0\\\\x_1=-2\in ODZ\ ,\ x_2=3\in ODZ\ \ \ (teorema\ Vieta)\\\\Proverka:\ x=-2:\ \sqrt{4}-\sqrt{9}=2-3=-1\ne 1\ \ ne\ podxodit\\\\x=3:\ \ \sqrt{9}-\sqrt{4}=3-2=1\ ,\ \ 1=1\\\\Otvet:\ \ x=3\ .$