Tìm tập xác định \({\rm{D}}\) của hàm số \(y = \dfrac{{3 - x}}{{\sqrt {4 - 3x - {x^2}} }}.\)

$\left( {3; + \,\infty } \right).$

\(\left( { - \,\infty ;3} \right).\)

\(\left( { - \,3;3} \right).\)

\(\mathbb{R}.\)

\({\left( {x + 3} \right)^2} + {\left( {y - 1} \right)^2} = \sqrt 5 .\)

\({\left( {x - 3} \right)^2} + {\left( {y + 1} \right)^2} = \sqrt 5 .\)

\({\left( {x - 3} \right)^2} + {\left( {y + 1} \right)^2} = 5.\)

\({\left( {x + 3} \right)^2} + {\left( {y - 1} \right)^2} = 5.\)

\({x^2} + 2{y^2} - 4x - 8y + 1 = 0\)

\(4{x^2} + {y^2} - 10x - 6y - 2 = 0\)

\({x^2} + {y^2} - 2x - 8y + 20 = 0\)

\({x^2} + {y^2} - 4x + 6y - 12 = 0\)

\(x \in \left( {\dfrac{2}{3};1} \right).\)

\(x \in \left( { - \infty ;\dfrac{2}{3}} \right) \cup \left( {1; + \infty } \right).\)

\(x \in \left( {\dfrac{2}{3};1} \right].\)

\(x \in \left( { - \infty ;1} \right) \cup \left( {\dfrac{2}{3}; + \infty } \right).\)

$S = \left( { - \,4; - \,1} \right).$

$S = \left( { - \,1; - \dfrac{1}{{11}}} \right).$

$S = \left( { - \,\infty ; - \dfrac{1}{{11}}} \right).$

$S = \left[ { - \dfrac{1}{{11}}; + \,\infty } \right).$

\(2{x^2} + 3y > 0.\)

\({x^2} + {y^2} < 2.\)

\(x + {y^2} \ge 0.\)

\(x + y \ge 0.\)

$\dfrac{{{x^2}}}{{100}} + \dfrac{{{y^2}}}{{81}} = 1$.

$\dfrac{{{x^2}}}{{34}} + \dfrac{{{y^2}}}{{25}} = 1$.

$\dfrac{{{x^2}}}{{25}} + \dfrac{{{y^2}}}{9} = 1$

$\dfrac{{{x^2}}}{{25}} - \dfrac{{{y^2}}}{{16}} = 1$.