Question

розвязати нерівність √(х²-1)>х

Answer 1

$\sqrt{x^2-1}>x\\\\\sqrt{f(x)}>g(x)\; \; \Leftrightarrow \; \; \left \{ {{g(x)\geq 0,} \atop {f(x)>(g(x))^2}} \right.\; \; ili\; \; \left \{ {{g(x)<0,} \atop {f(x)\geq 0.}} \right.\\\\a)\left \{ {{x^2-1\geq 0} \atop {x^2-1>x^2}} \right. \; \left \{ {{(x-1)(x+1)\geq 0} \atop {-1>0}} \right.\; \; \Rigtarrow \; \; x\in \varnothing \\\\b)\; \; \left \{ {{x<0} \atop {x^2-1\geq 0}} \right. \; \left \{ {{x<0} \atop {(x-1)(x+1)\geq 0}} \right.\; \left \{ {{x\in (-\infty ,0)} \atop {x\in (-\infty ,-1]\cup [\, 1,+\infty )}} \right. \; \Rightarrow \; \; x\in (-\infty ,-1\, ]\\\\Otvet:\; \; x\in (-\infty ,-1\, ]\; .$