Question

Решите неравенство
(x^2+1)^2-3(x^2+1)<10

Answer 1

$(x^2+1)^2-3(x^2+1)<10\\\\t=x^2+1>0\ ,\ \ \ t^2-3t-10<0\ \ ,\\\\\star \ \ t^2-3t-10=0\ \ \to \ \ t_1=-2\ ,\ t_2=5\ \ \star \\\\(t+2)(t-5)<0\ \ ,\ \ \ znaki:\ +++(-2)---(5)+++\\\\-2<t<5\ \ \ \to \ \ \ -2<(x^2+1)<5\\\\t>0\ \ \ \to \ \ \ 0<(x^2+1)<5\ \ \ \to \ \ \ \left\{\begin{array}{ccc}x^2+1>0\\x^2+1<5\end{array}\right\ \ \ \to \\\\\\x^2-4<0\ \ ,\ \ (x-2)(x+2)<0\ \ ,\ \ \ +++(-2)---(2)+++\\\\\boxed {\ x\in (-2,2)\ }$