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Правильный ответ

1.
$log_{н}(x^2-3x+2)\ \textgreater \ log_{н}(6-x^2+4x)$

ОДЗ:
+ - +
{x²-3x+2>0 {(x-2)(x-1)>0 ------(1)----------(2)----------->
//////// /////////////
- + -
{6-x²+4x>0 {(x-2+√10)(x-2-√10)>0 ------(2-√10)------------(2+√10)------->
////////////////////////

Общее:
x∈(2-√10; 1)∪(2; 2+√10)

$x^{2} -3x+2=0 \\ D=9-8=1 \\ x_1= \frac{3+1}{2} =2 \\ x_2= \frac{3-1}{2} =1$

$-x^2+4x+6=0 \\ D=16-4*(-1)*6=16+24=40 \\ x_1= \frac{-4+2 \sqrt{10} }{-2}=2- \sqrt{10} \\ x_2= \frac{-4-2 \sqrt{10} }{-2}=2+ \sqrt{10}$

$x^{2} -3x+2\ \textless \ 6-x^2+4x \\ x^2-3x+2-6+x^2-4x\ \textless \ 0 \\2x^2-7x-4\ \textless \ 0 \\ D= 49-4*2*(-4)=81 \\ x_1= \frac{7+9}{4}=4 \\ x_2= \frac{7-9}{4}=-0.5$
+ - +
-------(-0.5)-----------------(4)-------------->
///////////////////////////
Ответ: (-0,5; 1)∪(2; 4)

2.
$log_{н} \frac{2x-1}{16-x^2} \geq 2$

ОДЗ:
$\frac{2x-1}{16-x^2}\ \textgreater \ 0 \\ \frac{2x-1}{(4-x)(4+x)}\ \textgreater \ 0$
+ - + -
--------(-4)----------(¹/₂)----------(4)--------->
//////////// /////////////////
x∈(-∞; -4)∪(¹/₂; 4)

$log_ \frac{1}{2} \frac{2x-1}{16-x^2} \ \textgreater \ log_ \frac{1}{2} \frac{1}{4} \\ \frac{2x-1}{16-x^2} \leq \frac{1}{4} \\ \frac{2x-1}{16-x^2} - \frac{1}{4} \leq 0 \\ \frac{8x-4-16+x^2}{4(4-x)(4+x)} \leq 0 \\ \frac{x^2+8x-20}{4(4-x)(4+x)} \leq 0 \\ x^2+8x-20=0 \\ D=64+80=144 \\ x_1= \frac{-8+12}{2}=2 \\ x_1= \frac{-8-12}{2}=-10 \\ \frac{(x-2)(x+10)}{4(4-x)(4+x)} \leq 0$
- + - + -
----------[-10]----------(-4)-----------[2]-----------(4)-------->
//////////////// /////////////////// //////////
Ответ: (-∞; -10]∪(¹/₂; 2]

3.
$lg^2x-lgx-2\ \textless \ 0$
ОДЗ: х>0

Замена: $lgx=a$
$a^2-a-2\ \textless \ 0 \\ a^2-a-2=0 \\ D=(-1)^2+8=9 \\ a_1= \frac{1+3}{2}=2 \\ a_2= \frac{1-3}{2}=-1 \\ \\ (a-2)(a+1)\ \textless \ 0$
+ - +
----------(-1)--------------(2)----------->
/////////////////////
$-1\ \textless \ lgx\ \textless \ 2 \\ \\ \left \{ {{lgx\ \textless \ 2} \atop {lgx\ \textgreater \ -1}} \right. \\ \left \{ {{lgx\ \textless \ lg100} \atop {lgx\ \textgreater \ lg \frac{1}{10} }} \right. \\ \left \{ {{x\ \textless \ 100} \atop {x\ \textgreater \ \frac{1}{10} }} \right.$

\\\\\\\\\\\\\\\XXXXXXXXX///////////////
---------(0.1)-------------(100)----------->

Ответ: (0,1; 100)

chamomile03_zn (29.4k баллов) 10 Май, 18