`Q=(1/(\sqrt{a}-1)-1/(\sqrt{a})):((\sqrt{a}+1)/(\sqrt{a}-2)-(\sqrt{a}+2)/(\sqrt{a}-1))`

Điều kiện xác định :

$\begin{cases}a\geq0\\\sqrt{a}-1\ne 0\\\sqrt{a}\ne 0\\ \sqrt{a}-2\ne 0\end{cases}$

`<=>` $\begin{cases}a\geq 0\\ \sqrt{a}\ne 1\\ \sqrt{a}\ne 0\\ \sqrt{a}\ne 2\end{cases}$

`<=>` $\begin{cases}a\geq 0\\a\ne 1\\a\ne 0\\ a\ne 4\end{cases}$

`<=>` $\begin{cases}a>0\\a\ne 1\\a\ne 4\end{cases}$

Khi đó :

`Q=((\sqrt{a})/(\sqrt{a}(\sqrt{a}-1))-(\sqrt{a}-1)/(\sqrt{a}(\sqrt{a}-1))):(((\sqrt{a}+1)(\sqrt{a}-1))/((\sqrt{a}-2)(\sqrt{a}-1))-((\sqrt{a}+2)(\sqrt{a}-2))/((\sqrt{a}-1)(\sqrt{a}-2)))`

`=(\sqrt{a}-(\sqrt{a}-1))/(\sqrt{a}(\sqrt{a}-1)):([(\sqrt{a})^2 -1^2 ]-[(\sqrt{a})^2 -2^2])/((\sqrt{a}-2)(sqrt{a}-1))`

`=(\sqrt{a}-\sqrt{a}+1)/(\sqrt{a}(\sqrt{a}-1)):((a-1)-(a-4))/((\sqrt{a}-2)(sqrt{a}-1))`

`=1/(\sqrt{a}(\sqrt{a}-1)):(a-1-a+4)/((\sqrt{a}-2)(\sqrt{a}-1))`

`=1/(\sqrt{a}(\sqrt{a}-1)). ((\sqrt{a}-2)(\sqrt{a}-1))/3`

`=((\sqrt{a}-2)(sqrt{a}-1))/(3\sqrt{a}(\sqrt{a}-1))`

`=(\sqrt{a}-2)/(3\sqrt{a})`

Vậy `Q=(\sqrt{a}-2)/(3\sqrt{a})` với `a>0,a\ne 1;4`

$\boxed{\bullet\text{ Áp dụng :}\\a^2 -b^2 =(a-b)(a+b)}$

ĐKXĐ:

`{(sqrta-1\ne0),(sqrta\ne0),(sqrta-2\ne0),(a\ge0):}`

`<=> {(sqrta\ne1),(a\ne0),(sqrta\ne2),(a\ge0):}`

`<=> {(a\ne1),(a>0),(a\ne4):}`

`Q = (1/(sqrta-1)-1/sqrta) : ((sqrta+1)/(sqrta-2)-(sqrta+2)/(sqrta-1))`

`= (sqrta-(sqrta-1))/(sqrta(sqrta-1)) : ((sqrta-1)(sqrta+1)-(sqrta-2)(sqrta+2))/((sqrta-2)(sqrta+1))`

`= (sqrta-sqrta+1)/(sqrta(sqrta-1)) : ((sqrta)^2-1^2-[(sqrta)^2-2^2])/((sqrta-2)(sqrta+1))`

`= 1/(sqrta(sqrta-1)) : (a-1-a+4)/((sqrta-2)(sqrta+1))`

`= 1/(sqrta(sqrta-1)) * ((sqrta-2)(sqrta+1))/3`

`= (sqrta-2)/(3sqrta)`