tex]...

Sherlock444 @Sherlock444

August 2021 1 16 Report

[tex]x+ \sqrt{x+18} \leq 2[/tex]

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Verified answer

$x+ \sqrt{x+18} \leq 2$
$\sqrt{x+18} \leq 2-x$
$\left \{ {{x+18 \geq 0} \atop {2-x \geq 0}} \atop {( \sqrt{x+18})^2 \leq (2-x)^2}} \right.
$
$\left \{ {{x \geq -18} \atop {x \leq 2}} \atop {x+18 \leq4+x^2-4x}} \right.
$
$\left \{ {{x \geq -18} \atop {x \leq 2}} \atop {-x^2+5x+14 \leq0}} \right.
$
$\left \{ {{x \geq -18} \atop {x \leq 2}} \atop {x^2-5x-14 \geq 0}} \right. 
$
$x^2-5x-14=0$
$D=(-5)^2-4*1*(-14)=25+56=81$
$x_1= \frac{5+9}{2}=7$
$x_1= \frac{5-9}{2}=-2$

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Ответ: [ -18; -2]

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