Tập nghiệm của bất phương trình $\left| {{x^2} - 5x + 4} \right| \le {x^2} + 6x + 5$ là:

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

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

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

\(\mathbb{R}.\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(x + y \ge 0.\)

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

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

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

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