đk a > 0 ; a ≠ 1
\(Q=\left(\dfrac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{a-1}\right).\left(\dfrac{a-1}{\sqrt{a}}\right)\\ =\left(\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{a-1}\right).\left(\dfrac{a-1}{\sqrt{a}}\right)\\ =\dfrac{-4\sqrt{a}}{1}.\dfrac{1}{\sqrt{a}}\\ =-4\)
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