Пожалуйста, помогите решить номер 6.25

Question 1

Question 2

$6.25.$

$\sqrt{ x^{2} -3x} \ \textless \ 5-x$

$\left \{ {{ x^{2} -3x \geq 0} \atop {5-x\ \textgreater \ 0}} \atop {(\sqrt{ x^{2} -3x})^2 \ \textless \ (5-x)^2}} \right.$

$\left \{ {{ x(x-3) \geq 0} \atop {-x\ \textgreater \ -5}} \atop { x^{2} -3x} \ \textless \ 25-10x+x^2}} \right.$

$\left \{ {{ x(x-3) \geq 0} \atop {x\ \textless \ 5}} \atop { x^{2} -3x}-25+10x-x^2\ \textless \ 0}} \right.$

$\left \{ {{ x(x-3) \geq 0} \atop {x\ \textless \ 5}} \atop { 7x-25\ \textless \ 0}} \right.$

$\left \{ {{ x(x-3) \geq 0} \atop {x\ \textless \ 5}} \atop { 7x\ \textless \ 25}} \right.$

$\left \{ {{ x(x-3) \geq 0} \atop {x\ \textless \ 5}} \atop { x\ \textless \ 3 \frac{4}{7} }} \right.$

+ - +
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$x$ ∈ $(-$ ∞ $;0]$ ∪ $[3;3 \frac{4}{7})$

Ответ: $(-$ ∞ $;0]$ ∪ $[3;3 \frac{4}{7})$

mukus13_zn · Answer 1 · 2018-05-14T11:13:39+0000

$6.25.$

$\sqrt{ x^{2} -3x} \ \textless \ 5-x$

$\left \{ {{ x^{2} -3x \geq 0} \atop {5-x\ \textgreater \ 0}} \atop {(\sqrt{ x^{2} -3x})^2 \ \textless \ (5-x)^2}} \right.$

$\left \{ {{ x(x-3) \geq 0} \atop {-x\ \textgreater \ -5}} \atop { x^{2} -3x} \ \textless \ 25-10x+x^2}} \right.$

$\left \{ {{ x(x-3) \geq 0} \atop {x\ \textless \ 5}} \atop { x^{2} -3x}-25+10x-x^2\ \textless \ 0}} \right.$

$\left \{ {{ x(x-3) \geq 0} \atop {x\ \textless \ 5}} \atop { 7x-25\ \textless \ 0}} \right.$

$\left \{ {{ x(x-3) \geq 0} \atop {x\ \textless \ 5}} \atop { 7x\ \textless \ 25}} \right.$

$\left \{ {{ x(x-3) \geq 0} \atop {x\ \textless \ 5}} \atop { x\ \textless \ 3 \frac{4}{7} }} \right.$

+ - +
---------------[0]-------------[3]--------------------
///////////////// ///////////////////////
--------------------------------------------(5)--------
///////////////////////////////////////////////
----------------------------------(3 4/7)-------------
////////////////////////////////////////

$x$ ∈ $(-$ ∞ $;0]$ ∪ $[3;3 \frac{4}{7})$

Ответ: $(-$ ∞ $;0]$ ∪ $[3;3 \frac{4}{7})$