[tex] \sqrt{x-5}+ \sqrt{10-x}\ \textless \ 3[/tex] . Created by Sherlock444. algebra-ru

ОДЗ: ∈ -----+-----(6)------ - ----------(9)------+-------////////////// ///////////////------[5]-------------------------------[10]-------- ////////////////////////////////////Ответ: [5;6) ∪

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Sherlock444 @Sherlock444

August 2021 1 13 Report

[tex] \sqrt{x-5}+ \sqrt{10-x}\ \textless \ 3[/tex]

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$\sqrt{x-5}+ \sqrt{10-x}\ \textless \ 3$
ОДЗ:
$\left \{ {{x-5 \geq 0} \atop {10-x \geq 0}} \right.$
$\left \{ {{x \geq 5} \atop {x \leq 10}} \right.$
$x$ ∈ $[5;10]$

$(\sqrt{x-5}+ \sqrt{10-x})^2\ \textless \ 3^2$
${x-5}+10-x}+2 \sqrt{(x-5)(10-x)} \ \textless \ 9$
$5+2 \sqrt{(x-5)(10-x)} \ \textless \ 9$
$2 \sqrt{(x-5)(10-x)} \ \textless \ 4$
$\sqrt{(x-5)(10-x)} \ \textless \ 2$
$( \sqrt{(x-5)(10-x)})^2 \ \textless \ 2^2$
$(x-5)(10-x)} \ \textless \ 4$
$10x- x^{2} -50+5x-4} \ \textless \ 0$
$- x^{2}+15x-54} \ \textless \ 0$
$x^{2}-15x+54} \ \textgreater \ 0$
$D=(-15)^2-4*1*54=9$
$x_1= \frac{15+3}{2}=9$
$x_2= \frac{15-3}{2}=6$

-----+-----(6)------ - ----------(9)------+-------
////////////// ///////////////
------[5]-------------------------------[10]--------
////////////////////////////////////

Ответ: [5;6) ∪ $(9;10]$

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