Решите срочно пожалуйста с решением огромное спасибо пойму любой подчерк

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Правильный ответ

$a)$

$\left \{ {{|x-2| \leq 5} \atop { \frac{x-7}{x+5} \geq 0 }} \right.$

$\left \{ {{-5 \leq x-2 \leq 5} \atop { \frac{x-7}{x+5} \geq 0 }} \right.$

$\left \{ {{-3 \leq x \leq 7} \atop { \frac{x-7}{x+5} \geq 0 }} \right.$

------------[-3]----------[7]------------
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+ - +
-----(-5)----------------[7]-------------
//////// ////////////////

Ответ: { $7$ }

$b)$

$\left \{ {{ \frac{x+1}{x-4} \geq 0 } \atop {2(x-2)\ \textgreater \ 3(x-3)}} \right.$

$\left \{ {{ \frac{x+1}{x-4} \geq 0 } \atop {2x-4\ \textgreater \ 3x-9}} \right.$

$\left \{ {{ \frac{x+1}{x-4} \geq 0 } \atop {2x-3x\ \textgreater \ -9+4}} \right.$

$\left \{ {{ \frac{x+1}{x-4} \geq 0 } \atop {-x\ \textgreater \ -5}} \right.$

$\left \{ {{ \frac{x+1}{x-4} \geq 0 } \atop {x\ \textless \ 5}} \right.$

+ - +
----------[-1]----------(4)------------
//////////// ///////////////

-------------------------------(5)-------
/////////////////////////////////

Ответ: $(-$ ∞ $;-1]$ ∪ $(4;5)$

$c)$

$\left \{ {{ \frac{x+5}{x-3} \geq 0 } \atop {12 \leq x^{2} +x}} \right.$

$\left \{ {{ \frac{x+5}{x-3} \geq 0 } \atop {12 - x^{2} -x \leq 0}} \right.$

$\left \{ {{ \frac{x+5}{x-3} \geq 0 } \atop {x^{2} +x -12 \geq 0}} \right.$

$\left \{ {{ \frac{x+5}{x-3} \geq 0 } \atop {(x-3)(x+4) \geq 0}} \right.$

$x^{2} +x-12=0$

$D=1+48=49$

$x_1= \frac{-1+7}{2}=3$

$x_2= \frac{-1-7}{2}=-4$

$x^{2} +x-12=(x-3)(x+4)$

+ - +
---------[-5]-------------(3)------------
//////////// //////////////

+ - +
---------------[-4]-------[3]-----------
///////////////// //////////////

Ответ: $(-$ ∞ $;-5]$ ∪ $(3;+$ ∞ $)$

mukus13_zn (83.6k баллов) 30 Апр, 18