Question

Иррациональное уровнение

Answer 1

Ответ:

$\sqrt{2x-1}<x-2\ \ \ \Leftrightarrow \ \ \ \ \left\{\begin{array}{l}2x-1\geq 0\\x-2>0\\2x-1<(x-2)^2\end{array}\right\ \ \left\{\begin{array}{l}2x\geq 1\\x>2\\2x-1<x^2-4x+4\end{array}\right$

$\left\{\begin{array}{l}x\geq 0,5\\x>2\\x^2-6x+5>0\end{array}\right\ \ \left\{\begin{array}{l}x>2\\(x-1)(x-5)>0\end{array}\right\ \ \left\{\begin{array}{l}x>2\\x\in (-\infty ;\, 1)\cup (\, 5\, ;+\infty \, )\end{array}\right\\\\\\Otvet:\ \ x\in (\, 2\, ;\, 1\, )\cup (\, 5\, ;+\infty \, )\ .\\\\\\\star \ \ \ x^2-6x+5>0\ \ , \ \ x_1=1\ ,\ x_25\ \ \ (teorema\ Vieta)\\\\(x-1)(x-5)>0\ \ \ \ +++(1)---(5)+++\\\\x\in (-\infty ;\, 1)\cup (\, 5\, ;+\infty \, )\ \ \ \star$