$f(x)=\sqrt{2x-3}-\sqrt{7-x}$
$f'(x)=(\sqrt{2x-3}-\sqrt{7-x})'$
$f'(x)=\frac {1}{2\sqrt{2x-3}}(2x-3)'-\frac {1}{2\sqrt{7-x}}(7-x)'$
$f'(x)=\frac {2}{2\sqrt{2x-3}}+\frac {1}{2\sqrt{7-x}}$
Để $f'(x)\leq 0$ thì $f'(x)=\frac {2}{2\sqrt{2x-3}}+\frac {1}{2\sqrt{7-x}}\leq 0$
=> $.......$