Question

2,3,4 номера пожалуйста ......

Answer 1

Ответ:

$2)\ \ log_{0,2}(x-2)+log_{0,2}x\geq log_{0,2}(2x-3)\ \ ,\ \ \ ODZ:\ x>2\ ,\\\\log_{0,2}(x^2-2x)\geq log_0,2}(2x-3)\\\\0<0,2<1\ \ \to \ \ \ x^2-2x\leq 2x-3\ \ ,\ \ \ x^2-4x+3\leq 0\ \ ,\\\\(x-1)(x-3)\leq 0\ \ ,\ \ \ \ +++(1)---(3)+++\\\\x\in (\ 1\ ;\ 3\ )\ +\ ODZ\ \ \Rightarrow \ \ \ \boxed{\ x\in (\ 2\ ;\ 3\ )\ }$

$3)\ \ \dfrac{\sqrt{36-x^2}\cdot log_{0,5}\, x}{x-2}\leq 0\ \ ,\\\\ODZ:\ \left\{\begin{array}{l}x>0\\36-x^2\geq 0\end{array}\right\ \left\{\begin{array}{l}x>0\\(6-x)(6+x)\geq 0\end{array}\right\ \left\{\begin{array}{l}x>0\\x\in [-6\, ;\ 6\ ]\end{array}\right\ x\in (\ 0\ ;6\ ]\\\\\\Tak\ kak\ \ \sqrt{36-x^2}\geq 0\ \ ,\ \ i\ \ \sqrt{36-x^2}=0\ \ pri\ \ x=\pm 6\ ,\ to$

$a)\ \ \left\{\begin{array}{l}log_{0,5}\, x\leq 0\\x-2>0\end{array}\right\ \ \left\{\begin{array}{l}x\geq 1\\x>2\end{array}\right\ \ \ \Rightarrow \ \ x\in (\ 2\ ;+\infty \, )\\\\\\\left\{\begin{array}{l}x\in (\ 0\ ;\ 6\ ]\\x\in (\ 2\ ;+\infty \, )\end{array}\right\ \ \ \Rightarrow \ \ \ \underline {\ x\in (\ 2\ ;\ 6\ ]\ }$

$b)\ \ \left\{\begin{array}{l}log_{0,5}\, x\geq 0\\x-2<0\end{array}\right\ \ \left\{\begin{array}{l}x\leq 1\\x<2\end{array}\right\ \ \ \Rightarrow \ \ x\in (-\infty \, ;\ 1\ ]\\\\\\\left\{\begin{array}{l}x\in (\ 0\ ;\ 6\ ]\\x\in [\ -\infty \, ;\ 1\ ]\\x=\pm 6\end{array}\right\ \ \ \Rightarrow \ \ \ \underline {\ x\in (\ 0\ ;\ 1\ ]\ }\\\\\\Otvet:\ \ x\in (\ 0\ ;1\ ]\cup (\ 2\ ;\ 6\ ]\ .$

$4)\ \ \left\{\begin{array}{l}3\sqrt{x+y}-2\sqrt{x-y}=4\\2\sqrt{x+y}-\sqrt{x-y}=3\ |\cdot (-2)\end{array}\right\ \oplus \ \left\{\begin{array}{l}-\sqrt{x+y}=-2\\2\sqrt{x+y}-\sqrt{x-y}=3 \end{array}\right\\\\\\\left\{\begin{array}{l}\sqrt{x+y}=2\\4-\sqrt{x-y}=3\end{array}\right\ \ \left\{\begin{array}{l}\sqrt{x+y}=2\\\sqrt{x-y}=1 \end{array}\right\ \ \left\{\begin{array}{l}x+y=4\\x-y=1\end{array}\right\ \oplus\ \ominus$

$\left\{\begin{array}{l}2x=5\\2y=3\end{array}\right\ \ \left\{\begin{array}{l}x=2,5\\y=1,5\end{array}\right\\\\\\Otvet:\ (\ 2,5\ ;\ 1,5)\ .$