Bài `5:`

`a)` Ta có: $\begin{cases} x\ge0\\x\ne0\\\sqrt{x}\ne1\\\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}\ne0 \end{cases}$

$\begin{cases} x\ge0\\x\ne0\\\sqrt{x}\ne1\\\dfrac{(\sqrt{x}+1)-(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)}\ne0 \end{cases}$

$\begin{cases} x>0\\x\ne1\\\dfrac{2}{x-1}\ne0\rm{(lđ)} \end{cases}$

Vậy điều kiện xác định của biểu thức `P` là `x>0` và `x\ne1`

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`b) P=(1/(\sqrtx-1)-1/(\sqrtx)):(1/(\sqrtx-1)-1/(\sqrtx+1)) (x>0;x\ne1)`

`P=((\sqrtx)/(\sqrtx(\sqrtx-1))-(\sqrtx-1)/(\sqrtx(\sqrtx-1))):((\sqrtx+1)/((\sqrtx-1)(\sqrtx+1))-(\sqrtx-1)/((\sqrtx-1)(\sqrtx+1)))`

`P=(\sqrtx-(\sqrtx-1))/(\sqrtx(\sqrtx-1)):((\sqrtx+1)-(\sqrtx-1))/((\sqrtx-1)(\sqrtx+1))`

`P=(\sqrtx-\sqrtx+1)/(\sqrtx(\sqrtx-1)):(\sqrtx+1-\sqrtx+1)/((\sqrtx-1)(\sqrtx+1))`

`P=1/(\sqrtx(\sqrtx-1)):2/((\sqrtx-1)(\sqrtx+1))`

`P=1/(\sqrtx(\sqrtx-1))*((\sqrtx-1)(\sqrtx+1))/2`

`P=((\sqrtx-1)(\sqrtx+1))/(2\sqrtx(\sqrtx-1))`

`P=(\sqrtx+1)/(2\sqrtx)`

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`c)` Thay `x=4` vào `P`, ta được:

`P=(\sqrt4 +1)/(2\sqrt4)=(2+1)/(2.2)=3/4`

Vậy `P=3/4`