Алгебра: Решите неравенство: 2^(x/(x+((5x+3)/(x+1))+8≤2^((2x)/(x+1))

ОДЗ: Замена: + - +-----------[-8/9]------------[1]-------------///////////// /////////////---------------------(0)--------------------- ////////////////////// + - +--------------(-1)------------[0]------------------///////////////// ////////////////////ответ: ∞ ∪ ∞ ...

Решите неравенство: 2^(x/(x+((5x+3)/(x+1))+8≤2^((2x)/(x+1))

Ответы

shootdm100 04.10.2020 16:11

$2^{ \frac{x}{x+1} }- 2^{ \frac{5x+3}{x+1} }+8 \leq 2^{ \frac{2x}{x+1}$

ОДЗ:
$x+1 \neq 0$
$x \neq -1$

$2^{ \frac{x}{x+1} }- 2^{ \frac{2x}{x+1} } \leq 2^{ \frac{5x+3}{x+1} }-8$

$2^{ \frac{x}{x+1} }- (2^{ \frac{x}{x+1}})^2 } \leq 2^{ \frac{3x+3}{x+1} }* 2^{ \frac{2x}{x+1} } -8$

$2^{ \frac{x}{x+1} }- (2^{ \frac{x}{x+1}})^2 } \leq 2^{ \frac{3(x+1)}{x+1} }* (2^{ \frac{x}{x+1}} )^{2} } -8$

$2^{ \frac{x}{x+1} }- (2^{ \frac{x}{x+1}})^2 } \leq 2^{3}* (2^{ \frac{x}{x+1}} )^{2} } -8$

$2^{ \frac{x}{x+1} }- (2^{ \frac{x}{x+1}})^2 } \leq 8* (2^{ \frac{x}{x+1}} )^{2} } -8$

Замена:
$2^{ \frac{x}{x+1} }=t,$ $t\ \textgreater \ 0$

$t-t^2 \leq 8t^2-8$

$t-t^2-8t^2+8 \leq 0$

$-9t^2+t+8 \leq 0$

$9t^2-t-8 \geq 0$

D=(-1)^2-4*9*(-8)=289

$t_1= \frac{1+17}{18}=1$

$t_1= \frac{1-17}{18}=- \frac{8}{9}$

+ - +
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$t \geq 1$

$2^{ \frac{x}{x+1} } \geq 1$

$2^{ \frac{x}{x+1} } \geq2^0$

${ \frac{x}{x+1} } \geq0$

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ответ: