Question

Алгебра 11 класс. 3 примера даю 20 баллов ​

Answer 1

Ответ:

$5)\ \ log_{0,2}(2x+5)-log_{0,2}(16-x^2)<sin\dfrac{\pi }{2}\\\\ODZ:\ \left\{\begin{array}{ccc}2x+5>0\\16-x^2>0\end{array}\right\ \ \left\{\begin{array}{ccc}x>-2,5\\x\in (-4;4)\end{array}\right\ \ \to \ \ x\in (-2,5\, ;\, 4\, )\\\\\\log_{0,2}\dfrac{2x+5}{16-x^2}<1\ \ ,\ \ \ \dfrac{2x+5}{16-x^2}>0,2\ \ ,\ \ \ \dfrac{2x+5-3,2+0,2x^2}{(4-x)(4+x)}>0\ ,\\\\\\\dfrac{0,2x^2+2x+1,8}{(4-x)(4+x)}>0\ \ ,\ \ \dfrac{(x+9)(x+1)}{(x-4)(x+4)}<0\\\\\\znaki:\ \ \ +++(-9)---(-4)+++(-1)---(4)+++\\\\x\in (-9;-4)\cup (-1;\, 4\, )$

$\left\{\begin{array}{l}x\in (-9;-4)\cup (-1;\, 4\, )\\x\in (-2,5\, ;\, 4\, )\end{array}\right\ \ \ \Rightarrow \ \ \ x\in (-1\, ;\, 4\, )$

$6)\ \ lg(x^2-4x+13)>1\ \ ,\ \ ODZ:\ \ x^2-4x+13>0\ \to \ x\in R\\\\x^2-4x+13>10\ \ ,\ \ x^2-4x+3>0\ \ ,\ \ x_1=1\ ,\ x_2=3\\\\(x-1)(x-3)>0\ \ \ \ \ +++(1)---(3)+++\\\\x\in (-\infty ;1)\cup (3;+\infty )$

$7)\ \ log_{0,5}(1+x-\sqrt{x^2-4})<0\ \ ,\\\\ODZ:\ \left\{\begin{array}{l}1+x-\sqrt{x^2-4}>0\\x^2-4\geq 0\end{array}\right\ \ \left\{\begin{array}{l}\sqrt{x^2-4}<x+1\\(x-2)(x+2)\geq 0\end{array}\right\\\\\\\left\{\begin{array}{l}x^2-4<x^2+2x+1\\x+1>0\\x\in (-\infty ;-2\, ]\cup [\, 2\, ;+\infty )\end{array}\right\ \ \left\{\begin{array}{l}x>-2,5\\x>-1\\x\in (-\infty ;-2\, ]\cup [\, 2\, ;+\infty )\end{array}\right\\\\\\x\in [\, 2\, ;+\infty )$

$1+x-\sqrt{x^2-4}>1\ \ \to \ \ \ \sqrt{x^2-4}<x\ \ \to \ \ \ \left\{\begin{array}{l}x>0\\x^2-4<x^2\\x\in [\ 2\, ;+\infty )\end{array}\right\\\\\\\left\{\begin{array}{l}x>0\\-4<0\ (verno)\\x\in [\ 2\, ;+\infty )\end{array}\right\ \ \ \Rightarrow \ \ \ x\in [\ 2\, ;+\infty \, )$