https://lh4.googleusercontent.com/KAB_a1OoU3WuhujMhSy-IAcgkkwRbER-rwXXxqdfxyTMieIEYa4cx4QoAc9vZ1ahmpHzYV_EJwQcXfOFpfy3AfXPhGhzCzREeCF_6kmUXeXeKDfxrYSm6G08JulGrA8g4g=w740

Đáp án+giải thích các bước giải:

Ta có:

`T=\frac{\sqrt{x}}{\sqrt{x}-5}-\frac{5}{\sqrt{x}+5}-\frac{10\sqrt{x}}{x-25}` (ĐKXĐ: `x\ne25`)

`=\frac{\sqrt{x}.(\sqrt{x}+5)}{(sqrt{x}-5)(\sqrt{x}+5)}-\frac{5(\sqrt{x}-5)}{(\sqrt{x}-5)(\sqrt{x}+5)}`

`-\frac{10\sqrt{x}}{(\sqrt{x}-5)(\sqrt{x}+5)}`

`=\frac{x+5\sqrt{x}-5\sqrt{x}+25-10\sqrt{x}}{(\sqrt{x}-5)(\sqrt{x}+5)}`

`=\frac{x-10\sqrt{x}+25}{(\sqrt{x}-5)(\sqrt{x}+5)}`

`=\frac{(\sqrt{x}-5)^2}{(\sqrt{x}-5)(\sqrt{x}+5)}`

`=\frac{\sqrt{x}-5}{\sqrt{x}+5}`

Với `x\ne25`

Để T ∈ Z

`⇒\frac{\sqrt{x}-5}{\sqrt{x}+5}∈Z`

Ta có:

`\frac{\sqrt{x}-5}{\sqrt{x}+5}=1-\frac{10}{\sqrt{x}+5}`

`⇒T∈Z⇔\frac{10}{\sqrt{x}+5}∈Z`

`⇒(\sqrt{x}+5)∈Ư(10)={±1;±2;±5;±10}`

`\sqrt{x}+5=1⇒x=∅(l)`

`\sqrt{x}+5=-1⇒x=∅(l)`

`\sqrt{x}+5=2⇒x=∅(l)`

`\sqrt{x}+5=-2⇒x=∅(l)`

`\sqrt{x}+5=5⇒x=0(tmđk)`

`\sqrt{x}+5=-5⇒x=∅(l)`

`\sqrt{x}+5=10⇒x=25(ktmđk)`

`\sqrt{x}+5=-10⇒x=∅(l)`

Vậy có 1 giá trị của x để T ∈ Z

⇒ Chọn A