`a)`Thế `x=16` vào `A=\frac{\sqrt{x}+2}{\sqrt{x}-5}`, ta có:
`A=\frac{\sqrt{16}+2}{\sqrt{16}-5}`
`=\frac{4+2}{4-5}`
`=\frac{6}{-1}`
`=-6`
Vậy `A=-6` khi `x=16`
$c)B=\dfrac{3}{\sqrt{x}+5}+\dfrac{20-2\sqrt{x}}{x-25}$(sửa đề)
$=\dfrac{3(\sqrt{x}-5)}{(\sqrt{x}-5)(\sqrt{x}+5)}+\dfrac{20-2\sqrt{x}}{(\sqrt{x}-5)(\sqrt{x}+5)}$
$=\dfrac{3\sqrt{x}-15+20-2\sqrt{x}}{(\sqrt{x}-5)(\sqrt{x}+5)}$
$=\dfrac{\sqrt{x}+5}{(\sqrt{x}+5)(\sqrt{x}-5)}$
$=\dfrac{1}{\sqrt{x}-5}$
Vậy $B=\dfrac{1}{\sqrt{x}-5}$(đpcm)
