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

[latex] \sqrt{x-5}+ \sqrt{10-x}\ \textless \ 3 [/latex] ОДЗ: [latex] \left \{ {{x-5 \geq 0} \atop {10-x \geq 0}} \right. [/latex] [latex] \left \{ {{x \geq 5} \atop {x \leq 10}} \right. [/latex] [latex]x[/latex] ∈ [latex][5;10][/latex] [latex] (\sqrt{x-5}+ \sqrt{10-x})^2\ \textless \ 3^2 [/latex] [latex]{x-5}+10-x}+2 \sqrt{(x-5)(10-x)} \ \textless \ 9[/latex] [latex]5+2 \sqrt{(x-5)(10-x)} \ \textless \ 9[/latex] [latex]2 \sqrt{(x-5)(10-x)} \ \textless \ 4[/latex] [latex] \sqrt{(x-5)(10-x)} \ \textless \ 2[/latex] [latex]( \sqrt{(x-5)(10-x)})^2 \ \textless \ 2^2[/latex] [latex](x-5)(10-x)} \ \textless \ 4[/latex] [latex]10x- x^{2} -50+5x-4} \ \textless \ 0[/latex] [latex]- x^{2}+15x-54} \ \textless \ 0[/latex] [latex]x^{2}-15x+54} \ \textgreater \ 0[/latex] [latex]D=(-15)^2-4*1*54=9[/latex] [latex]x_1= \frac{15+3}{2}=9 [/latex] [latex]x_2= \frac{15-3}{2}=6[/latex] -----+-----(6)------ - ----------(9)------+------- //////////////                               /////////////// ------[5]-------------------------------[10]--------            //////////////////////////////////// Ответ: [5;6) ∪ [latex](9;10][/latex]