Question

$\sqrt{x^2-9} = x^2-21$
помогите пожалуйста!!

Answer 1

$\displaystyle \sqrt{x^2-9}=x^2-21\\x^2-9=x^4-42x^2+441\\x^2-9-x^4+42x^2-441=0\\43x^2-450-x^4=0\\-x^4+43x^2-450=0\\-t^2+43t-450=0\\t=18,t=25\\x^2=18,x^2=25\\x=-3\sqrt{2},x=3\sqrt{2},x=-5,x=5\\\sqrt{(-3\sqrt{2})^2-9 }=(-3\sqrt{2})^2-21, \sqrt{(3\sqrt{2})^2-9 }=(3\sqrt{2})^2-21,\\ \\ \sqrt{(-5)^2-9}=(-5)^2-21 ,\sqrt{5^2-9}=5^2-21\\\\ 3=-3, 3=-3,4=4,4=4\\x\neq -3\sqrt{2},x\neq 3\sqrt{2},x=-5,x=5\\x_{1}=-5,x_{2}=5$