Вопрос в картинках...

Question 1

Решите задачу:

$\frac{3}{(2^{2-x^2} -1)^2} - \frac{4}{2^{2-x^2}-1} +1 \geq 0$

Question 2

$\frac{3}{( 2^{2- x^{2} }-1 ) ^{2} } - \frac{4}{ 2^{2- x^{2} }-1 } +1 \geq 0$
ОДЗ:
$2^{2- x^{2} } -1 \neq 0 2^{2- x^{2} } \neq 1 2^{2- x^{2} } \neq 2^{0}$
2-x²≠0. x≠+-√2

замена переменной:
$2^{2- x^{2} } -1=t, t\ \textgreater \ 0$
$\frac{3}{ t^{2} } - \frac{4}{t} +1 \leq 0, \frac{3-4t+ t^{2} }{ t^{2} } \leq 0 \left \{ {{t ^{2}\ \textgreater \ 0 } \atop { t^{2} -4t+3 \leq 0}} \right.$
t²-4t+3=0. t₁=3, t₂=1
+ - +
------------[1]----------[3]------------>t

t≥1. t≤3

t²>0. t<0, t>0
/ / / / / / / / / / / / / / / / / / / / / / / / / / /
------------(0)--------[1]---------[3]-------------------->t
\ \ \ \ \ \ \ \
обратная замена:

t≥1.
$2^{2- x^{2} }-1 \geq 1 2^{2- x^{2} } \geq 2^{1} 2- x^{2} \geq 1$
(1-x)*(1+x)≥0
-1≤x≤1

t≤3
$2^{2- x^{2} } -1 \leq 3 2^{2- x^{2} } \leq 2^{2}$
2-x²≤2, -x²≤0 нет решений.
x∈[-1;1]

kirichekov_zn · Answer 1 · 2018-04-25T15:23:54+0000

$\frac{3}{( 2^{2- x^{2} }-1 ) ^{2} } - \frac{4}{ 2^{2- x^{2} }-1 } +1 \geq 0$
ОДЗ:
$2^{2- x^{2} } -1 \neq 0 2^{2- x^{2} } \neq 1 2^{2- x^{2} } \neq 2^{0}$
2-x²≠0. x≠+-√2

замена переменной:
$2^{2- x^{2} } -1=t, t\ \textgreater \ 0$
$\frac{3}{ t^{2} } - \frac{4}{t} +1 \leq 0, \frac{3-4t+ t^{2} }{ t^{2} } \leq 0 \left \{ {{t ^{2}\ \textgreater \ 0 } \atop { t^{2} -4t+3 \leq 0}} \right.$
t²-4t+3=0. t₁=3, t₂=1
+ - +
------------[1]----------[3]------------>t

t≥1. t≤3

t²>0. t<0, t>0
/ / / / / / / / / / / / / / / / / / / / / / / / / / /
------------(0)--------[1]---------[3]-------------------->t
\ \ \ \ \ \ \ \
обратная замена:

t≥1.
$2^{2- x^{2} }-1 \geq 1 2^{2- x^{2} } \geq 2^{1} 2- x^{2} \geq 1$
(1-x)*(1+x)≥0
-1≤x≤1

t≤3
$2^{2- x^{2} } -1 \leq 3 2^{2- x^{2} } \leq 2^{2}$
2-x²≤2, -x²≤0 нет решений.
x∈[-1;1]