Question

Решите неравенство: $x^{2} -2\sqrt{x^{2}-x } \ \textless \ x+15$

Answer 1

Ответ:

$x\in(\dfrac{1-\sqrt{101}}{2};0]\cup[1;\dfrac{1+\sqrt{101}}{2})$

Пошаговое объяснение:

$x^2-2\sqrt{x^2-x}<x+15\\ (x^2-x)-2\sqrt{x^2-x}+1<16\\ (\sqrt{x^2-x}-1)^2<4^2\\ (\sqrt{x^2-x}-5)\underbrace{(\sqrt{x^2-x}+3)}_{>\;0}<0\\ \sqrt{x^2-x}-5<0\\ \sqrt{x^2-x}<5\\ \left\{\begin{array}{c}x^2-x<25&x^2-x\geq 0\end{array}\right. =>\left\{\begin{array}{c}(x-\dfrac{1}{2})^2<\dfrac{101}{4}&x(x-1)\geq 0\end{array}\right. =>\left\{\begin{array}{c}\dfrac{1-\sqrt{101}}{2}<x<\dfrac{1+\sqrt{101}}{2}&x\in(-\infty;0]\cup[1;+\infty)\end{array}\right.=>x\in(\dfrac{1-\sqrt{101}}{2};0]\cup[1;\dfrac{1+\sqrt{101}}{2})$