Để bất phương trình \(\sqrt {(x + 5)(3 - x)} \le {x^2} + 2x + a\) nghiệm đúng \(\forall x \in \left[ { - 5;3} \right]\), tham số \(a\) phải thỏa điều kiện:

\(\overrightarrow {{n_1}}  = \left( {3;2} \right)\).

\(\overrightarrow {{n_2}}  = \left( { - 4; - 6} \right)\).

\(\overrightarrow {{n_3}}  = \left( {2; - 3} \right)\)

\(\overrightarrow {{n_4}} = \left( { - 2;3} \right)\).

$x \in \left( {0; + \infty } \right).$        

$x \in \left( { - 2; + \infty } \right).$

$\forall x \in \mathbb{R}.$

$x \in \left( { - \infty ;2} \right).$

\(\dfrac{1}{5}\)

\(\dfrac{{201}}{{1000}}\)      

\(\dfrac{{200}}{{999}}\)

\(\dfrac{{199}}{{999}}\)

$\left\{ {0; - 2} \right\}$

$\left\{ 0 \right\}$ 

$\left\{ { - 2} \right\}$

\(\emptyset \)

\(9\)

\(18\)

\(36\)

\(39\)

\(6\)     

\(5\)

\(1\)

\(7\)

$\left( { - \infty ;\dfrac{1}{2}} \right]$.

$\left[ {2; + \infty } \right)$.

$\left( { - \infty ;\dfrac{1}{2}} \right] \cup \left[ {2; + \infty } \right)$.

$\left[ {\dfrac{1}{2};2} \right]$.