Question

Решите иррациональное неравенство
помогите пожалуйста

Answer 1

$(x^2-2x)-sqrt{x^2-2x}-12 textless 0\ sqrt{x^2-2x} = t$
$begin{cases} t geq 0 \ t^2-t-12 textless 0end{cases} textless = textgreater begin{cases} t geq 0 \ (t+3)(t-4) textless 0end{cases} textless = textgreater begin{cases} t geq 0 \ -3 textless t textless 4end{cases}\ \ = textgreater 0 leq t textless 4$
$0 leq sqrt{x^2-2x} textless 4\ 0 leq x^2-2x textless 16\ begin{cases} x^2-2x geq 0 \ x^2-2x -16 textless 0 end{cases} textless = textgreater begin{cases} x(x-2) geq 0 \ (x-1+sqrt{17})(x-1-sqrt{17}) textless 0end{cases} textless = textgreater$
$begin{cases} x in (-infty; 0] cup [2; +infty) \ x in (1-sqrt{17}; 1+sqrt{17}) end{cases} = textgreater boxed {x in (1-sqrt{17};0] cup [2; 1+sqrt{17}) }$