$f(x) = \sqrt{x+2} - \sqrt{5-x} + \sqrt{(x+2)(5-x)}$
+ $D=[-2;5]$
+ $f'(x) = \dfrac{1}{2\sqrt{x+2}}+\dfrac{1}{2\sqrt{5-x}} + \dfrac{3-2x}{2\sqrt{(x+2)(5-x)}}$
Giải phương trình $f'(x) =0$
$\Leftrightarrow \dfrac{\sqrt{5-x}+\sqrt{x+2}+3-2x}{2\sqrt{(x+2)(5-x)}} =0$
$\Leftrightarrow \sqrt{5-x} +\sqrt{x+2} +3-2x=0$
$\Leftrightarrow 5-x+2\sqrt{(5-x)(x+2)} + x+2=(2x-3)^2=4x^2-12x+9$
$\Leftrightarrow 7+2\sqrt{-x^2+3x+10} - 4x^2 +12x-2=0$
$\Leftrightarrow 2\sqrt{-x^3+3x+10} - 4x^2+12x-2=0$
$\Leftrightarrow \sqrt{-x^2+3x+10} =4x^4-24x^3+40x^2 - 12x+1$
$\Leftrightarrow 4x^4 - 24x^3 +41x^2-15x-9=0$
$\Leftrightarrow (2x-3)(2x^3-9x^2+7x+3)=0$
$\Rightarrow \left[ \begin{array}{l}x=\dfrac{3}{2}\\x=\dfrac{3\pm \sqrt{13}}{2} \end{array} \right.$
BBT :
Vậy $\text{min} = - \sqrt{7}$ khi $x=-2$
$\text{Max} = 4$ khi $x= \dfrac{3+\sqrt{13}}{2}$
