Question

алгебра 1-3 желательно с решением

Answer 1

$1)\ \ y=\sqrt{x^3-5x^2+6x}\\\\OOF:\ x^3-5x^2+6x\geq 0\ \ ,\ \ \ x(x^2-5x+6)\geq 0\ \ ,\ \ x(x-2)(x-3)\geq 0\ ,\\\\znaki:\ \ ---[\, 0\, ]+++[\ 2\ ]---[\ 3\ ]+++\\\\x\in [\ 0\ ;2\ ]\cup [\ 3;+\infty )\\\\\\2)\ \ \sqrt{6x-x^2}+\dfrac{1}{x-5}\\\\\\ODZ:\ \left\{\begin{array}{l}6x-x^2\geq 0\\x-5\ne 0\end{array}\right\ \ \left\{\begin{array}{l}x(6-x)\geq 0\\x\ne 5\end{array}\right\ \ \left\{\begin{array}{l}x\in [\ 0\ ;\ 6\ ]\\x\ne 5\end{array}\right$

$x\in [\ 0\ ;\ 5\ )\cup (\ 5\ ;\ 6\ ]\\\\\star \ \ x\, (6-x)\geq 0\ \ ,\ \ x\, (x-6)\leq 0\ \ ,\ \ \ +++[\, 0\, ]---[\, 6\, ]+++\\\\x\in [\ 0\ ;\ 6\ ]\ \ \star$

$3)\ \ \sqrt{x^2+5}=x+2\ \ \ \Leftrightarrow \ \ \left\{\begin{array}{l}x+2\geq 0\\x^2+5=(x+2)^2\end{array}\right\\\\\\ \left\{\begin{array}{l}x\geq -2\\x^2+5=x^2+4x+4\end{array}\right\ \ \left\{\begin{array}{l}x\geq -2\\4x=1\end{array}\right\ \ \left\{\begin{array}{l}x\geq -2\\x=0,25\end{array}\right$

$Otvet:\ \ x=0,25\ .$