Найдите область определения функции y = sqr( 10 - x^2 - 3x ) / sqr( 1 - x )

[latex]y= \frac{ \sqrt{10- x^{2} -3x} }{ \sqrt{1-x} } [/latex] [latex] \left \{ {{10- x^{2} -3x \geq 0} \atop {1-x\ \textgreater \ 0}} \right. [/latex] [latex] \left \{ {{- x^{2} -3x+10 \geq 0} \atop {-x\ \textgreater \ -1}} \right. [/latex] [latex] \left \{ {{x^{2} +3x-10 \leq 0} \atop {x\ \textless \ 1}} \right. [/latex] [latex]D=9+40=49[/latex] [latex]x_1=2[/latex] [latex]x_2=-5[/latex] [latex] \left \{ {{(x-2)(x+5) \leq 0} \atop {x\ \textless \ 1}} \right. [/latex] [latex]x[/latex]∈ [latex][-5;2][/latex] [latex]x[/latex] ∈ [latex](-[/latex]∞[latex];1)[/latex] Ответ: [latex]D(y)=[-5;1)[/latex]