Giải pt sqrt(x+2)+sqrt(5-x)+sqrt(x+2)(5-x)=4

Đáp án: $x = \frac{{3 + 3\sqrt 5 }}{2}$

Giải thích các bước giải:

$\begin{array}{l}
ĐKxđ: - 2 \le x \le 5\\
\sqrt {x + 2}  + \sqrt {5 - x}  = a\left( {a > 0} \right)\\
 \Rightarrow {a^2} = x + 2 + 5 - x + 2\sqrt {x + 2} .\sqrt {5 - x} \\
 \Rightarrow {a^2} = 7 + 2\sqrt {\left( {x - 2} \right)\left( {5 - x} \right)} \\
 \Rightarrow \sqrt {\left( {x - 2} \right)\left( {5 - x} \right)}  = \frac{{{a^2} - 7}}{2}\\
 \Rightarrow pt:a + \frac{{{a^2} - 7}}{2} = 4\\
 \Rightarrow {a^2} + 2a - 7 - 8 = 0\\
 \Rightarrow {a^2} + 2a - 15 = 0\\
 \Rightarrow \left( {a - 3} \right)\left( {a + 5} \right) = 0\\
 \Rightarrow a = 3\left( {do\,:a > 0} \right)\\
 \Rightarrow \left\{ \begin{array}{l}
\sqrt {x + 2}  + \sqrt {5 - x}  = 3\\
\sqrt {\left( {x - 2} \right)\left( {5 - x} \right)}  = 1
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
\sqrt {x + 2}  = \frac{{3 - \sqrt 5 }}{2}\\
\sqrt {5 - x}  = \frac{{3 + \sqrt 5 }}{2}
\end{array} \right.\\
\left\{ \begin{array}{l}
\sqrt {x + 2}  = \frac{{3 + \sqrt 5 }}{2}\\
\sqrt {5 - x}  = \frac{{3 - \sqrt 5 }}{2}
\end{array} \right.
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 2 = \frac{{7 - 3\sqrt 5 }}{2}\\
5 - x = \frac{{7 + 3\sqrt 5 }}{2}
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 2 = \frac{{7 + 3\sqrt 5 }}{2}\\
5 - x = \frac{{7 - 3\sqrt 5 }}{2}
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = \frac{{1 - 3\sqrt 5 }}{2}\\
x = \frac{{3 - 3\sqrt 5 }}{2}
\end{array} \right.\left( {ktm} \right)\\
\left\{ \begin{array}{l}
x = \frac{{3 + 3\sqrt 5 }}{2}\\
x = \frac{{3 + 3\sqrt 5 }}{2}
\end{array} \right.
\end{array} \right. \Rightarrow x = \frac{{3 + 3\sqrt 5 }}{2}\left( {tm} \right)
\end{array}$