$D=[-2;5]$

Xét hàm số $f(x)=\sqrt{x+2}-\sqrt{5-x}+\sqrt{(x+2)(5-x)}$ với $x\in [-2;5]$

$f'(x)=\dfrac{1}{2\sqrt{x+2}}+\dfrac{1}{2\sqrt{5-x}}+\dfrac{(-x^2+3x+10)'}{2\sqrt{(x+2)(5-x)}}$

$=\dfrac{1}{2\sqrt{x+2}}+\dfrac{1}{2\sqrt{5-x}}-\dfrac{2x-3}{2\sqrt{(x+2)(5-x)}}$

$=\dfrac{\sqrt{5-x}+\sqrt{x+2}+3-2x}{2\sqrt{(5-x)(x+2)}}$

$y'=0\to \sqrt{5-x}+\sqrt{x+2}=2x-3$ (*)

ĐK: $x\ge \dfrac{3}{2}$

(*) $\to 7+2\sqrt{(5-x)(x+2)}=4x^2-12x+9$

$\to \sqrt{-x^2+3x+10}=2x^2-6x+1$

$\to -x^2+3x+10=(2x^2-6x)^2+2(2x^2-6x)+1= 4x^4-24x^3+36x^2+4x^2-12x+1$

$\to x=\dfrac{3+\sqrt{13}}{2}; x=\dfrac{3}{2}$ (TM điều kiện), loại $x=\dfrac{3-\sqrt{13}}{2}$

 Ta có BBT như hình.

Vậy:

$\max\limits_{[-2;5]}f(x)=f\Big(\dfrac{3+\sqrt{13}}{2}\Big)=4$

$\min\limits_{[-2;5]}f(x)=f(-2)=-\sqrt7$

$f(x) = \sqrt{x+2} - \sqrt{5-x} + \sqrt{(x+2)(5-x)}$

+ $D=[-2;5]$

+ $f'(x) = \dfrac{1}{2\sqrt{x+2}}+\dfrac{1}{2\sqrt{5-x}} + \dfrac{3-2x}{2\sqrt{(x+2)(5-x)}}$

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 $x= \dfrac{3+\sqrt{13}}{2}$