Question

иррациональное неравенство с одз​

Answer 1

Ответ:

$1)\ \ \sqrt{x+1}>2\ \ ,\ \ \ ODZ:\ x+1\geq 0\ \ ,\ \ x\geq -1\ ,\\\\x+1>4\ \ ,\ \ x>3\\\\Otvet:\ \ x\in (\, 3\, ;+\infty \, )\ .\\\\\\2)\ \ \sqrt{x^2+x-2}<2\ \ ,\ \ \ \ ODZ:\ x^2+x-2\geq 0\ ,\ (x+2)(x-1)\geq 0\ ,\\\\x\in (-\infty ;-2\, ]\cup [\ 1\ ;+\infty )\\\\x^2+x-2<4\ \ ,\ \ x^2+x-6<0\ \ ,\ \ (x+3)(x-2)<0\\\\znaki:\ \ +++(-3)---(2)+++\ \ \ ,\ \ \ \ x\in (-3;\ 2\ )\\\\\left\{\begin{array}{l}x\in (-\infty ;-2\, ]\cup [\ 1\ ;+\infty )\\x\in (-3\ ;\ 2\ )\end{array}\right\ \ \ \ \Rightarrow$

$Otvet:\ \ x\in (-3\ ;\ -2\ ]\cup [\ 1\ ;2\ )\ .$

$3)\ \ \sqrt{2x-1}>x-2\ \ \ \Leftrightarrow \ \ \ \left\{\begin{array}{l}x-2\geq 0\\2x-1>(x-2)^2\end{array}\right\ \ ili\ \ \left\{\begin{array}{l}x-2<0\\2x-1\geq 0\end{array}\right\\\\\\\left\{\begin{array}{l}x\geq 2\\2x-1>x^2-4x+4\end{array}\right\ \ \ ili\ \ \ \left\{\begin{array}{l}x<2\\2x\geq 1\end{array}\right$

$\left\{\begin{array}{l}x\geq 2\\x^2-6x+5<0\end{array}\right\ \ \ ili\ \ \ \left\{\begin{array}{l}x<2\\x\geq 0,5\end{array}\right\\\\\\\left\{\begin{array}{l}x\geq 2\\(x-1)(x-5)<0\end{array}\right\ \ \ ili\ \ \ \ 0,5\leq x<2$

$\left\{\begin{array}{l}x\geq 2\\x\in (\ 1\ ;\ 5\ )\end{array}\right\ \ \ ili\ \ \ \ x\in [\ 0,5\ ;\ 2\ )\\\\\\{}\ \ \ x\in [\ 2\ ;\ 5\ )\ \ \ \ \ \ ili\ \ \ \ \ \ \ x\in [\ 0,5\ ;\ 2\ )\\\\\\Otvet:\ \ x\in [\ 0,5\ ;\ 5\ )\ .$

Answer 2

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