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August 2022 1 16 Report

1. Решите неравенство:
а). [tex]( \frac{1}{3}) ^{ x^{2} +2x}<( \frac{1}{3}) ^{ 32-2x} [/tex]
б). [tex] log_{8}( x^{2} -4x+3) \geq 1 [/tex]
2. Постройте график функции [tex]y=log _{2} \frac{ x^{2} -16}{x-4} [/tex]

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arsenlevadniy

Verified answer

$(\frac{1}{3}) ^{x^{2}+2x}32-2x, \\ 
x^2+4x-32>0, \\ x^2+4x-32=0, \\ x_1=-8, x_2=4, \\ (x+8)(x-4)>0, \\ \left [ {{x4;}} \right. \\ x\in(-\infty;-8)\cup(4;+\infty);$

$log_{8}(x^{2} -4x+3) \geq 1, \\ \left \{ {{x^{2} -4x+3>0,} \atop {x^{2} -4x+3 \geq 8;}} \right. \\ x^{2} -4x+3=0, \\ x_1=1, x_2=3, \\ x^{2} -4x-5=0, \\ x_1=-1, x_2=5, \\ \left \{ {{(x-1)(x-3)>0,} \atop {(x+1)(x-5)\geq0;}} \right. \left \{ {{ \left[ {{x3,}} \right. } \atop { \left [ {{x \leq -1,} \atop {x \geq 5;}} \right. }} \right. \left [ {{x \leq -1,} \atop {x \geq 5;}} \right. \\ x\in(-\infty;-1]\cup[5;+\infty);$

$y=log _{2} \frac{x^{2}-16}{x-4}, \\ 
1) \frac{x^{2}-16}{x-4}>0, \\ 
\frac{(x+4)(x-4)}{x-4}>0, \\ x-4 \neq 0,x \neq 4, \\ 
(x+4)(x-4)^2>0, \\ 
x+4>0, \\
x>-4, \\
D_y=(-4;4)\cup(4;+\infty) \\
2) y=log _{2} (x+4), \\
x+4=2^y, \\
x=2^y-4, \\ 
y=2^x-4, \\ 
E_y=R; \\ 
3) f(-x)=log _{2} \frac{(-x)^{2}-16}{-x-4}=-log _{2} \frac{x^{2}-16}{x+4}, \\ f(-x) \neq f(x), f(-x) \neq -f(x);
$
$4)y=\log _{2} (x+4), \\ 
\log _{2} (x+4)\gtrless0, \\ 
x+4\gtrless1, \\
x\gtrless-3, \\ 
x0;
$

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