Đáp án:
`17`
Giải thích các bước giải:
TXĐ: $D = [ - 1 ; 9 ] \backslash \{ 0 \}$
Ta có
\[\begin{array}{*{20}{l}}
{\dfrac{{x - 3}}{{3\sqrt {x + 1} + x + 3}} \ge \dfrac{{2\sqrt {9 - x} }}{x} \Leftrightarrow \dfrac{{(x + 1) - 4}}{{x + 1 + 3(\sqrt {x + 1} ) + 2}} \ge \dfrac{{2\sqrt {9 - x} }}{{x + 1 - 1}}}\\
{ \Leftrightarrow \dfrac{{{{(\sqrt {x + 1} + 1)}^2} - 4}}{{(\sqrt {x + 1} + 1)(\sqrt {x + 1} + 2)}} \ge \dfrac{{2\sqrt {9 - x} }}{{(\sqrt {x + 1} + 1)(\sqrt {x + 1} - 1)}}}\\
{ \Leftrightarrow \dfrac{{(\sqrt {x + 1} - 2)(\sqrt {x + 1} + 2)}}{{(\sqrt {x + 1} + 1)(\sqrt {x + 1} + 2)}} \ge \dfrac{{2\sqrt {9 - x} }}{{(\sqrt {x + 1} + 1)(\sqrt {x + 1} - 1)}}}\\
{ \Leftrightarrow (\sqrt {x + 1} - 2) \ge \dfrac{{2\sqrt {9 - x} }}{{(\sqrt {x + 1} - 1)}}}
\end{array}\]
\[\begin{array}{*{20}{l}}
{ \Leftrightarrow (\sqrt {x + 1} - 2)(\sqrt {x + 1} - 1) \ge 2\sqrt {9 - x} }\\
{ \Leftrightarrow (\sqrt {x + 1} - 2)\sqrt {x + 1} + 2 - 2\sqrt {9 - x} \ge 0}\\
{ \Leftrightarrow \sqrt {x + 1} (\sqrt {x + 1} - 3) + 2(1 - \sqrt {9 - x} ) \ge 0}\\
{ \Leftrightarrow \dfrac{{\sqrt {x + 1} (x - 8)}}{{\sqrt {x + 1} + 3}} + \dfrac{{2(x - 8)}}{{1 + \sqrt {9 - x} }} \ge 0}\\
{ \Leftrightarrow (x - 8)\left( {\dfrac{{\sqrt {x + 1} }}{{\sqrt {x + 1} + 3}} + \dfrac{2}{{1 + \sqrt {9 - x} }}} \right) \ge 0}\\
{ \Leftrightarrow x \ge 8\left( {do\dfrac{{\sqrt {x + 1} }}{{\sqrt {x + 1} + 3}} + \dfrac{2}{{1 + \sqrt {9 - x} }} \ge 0\forall x \in (0;9]} \right)}
\end{array}\]
⇒ `x∈{8;9}`
tổng là: `8+9=17`
