Ta có: \(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
\(=\frac{\left(a-1\right)^2}{4a}.\frac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}=\frac{\left(a-1\right)^2}{4a}.\frac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{a-1}=\frac{\left(a-1\right)^2}{4a}.\frac{-4\sqrt{a}}{a-1}=\frac{1-a}{\sqrt{a}}\)
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ĐKXĐ : a > 0
Vậy ĐKXĐ là a > 0 .
Ta có :\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
=\(\left(\frac{\sqrt{a}.\sqrt{a}}{2\sqrt{a}}-\frac{1}{2\sqrt{a}}\right)^2\left(\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
=\(\left(\frac{\sqrt{a}^2}{2\sqrt{a}}-\frac{1}{2\sqrt{a}}\right)^2\left(\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}^2-1^2}-\frac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}^2-1^2}\right)\)
= \(\left(\frac{|a|}{2\sqrt{a}}-\frac{1}{2\sqrt{a}}\right)^2\left(\frac{\left(\sqrt{a}-1\right)^2}{|a|-1}-\frac{\left(\sqrt{a}+1\right)^2}{|a|-1}\right)\)
= \(\left(\frac{a}{2\sqrt{a}}-\frac{1}{2\sqrt{a}}\right)^2\left(\frac{\left(\sqrt{a}-1\right)^2}{a-1}-\frac{\left(\sqrt{a}+1\right)^2}{a-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)^2\left(\frac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{a-1}\right)\)
= \(\frac{\left(a-1\right)^2}{\left(2\sqrt{a}\right)^2}\left(\frac{\left(\sqrt{a}-1+\sqrt{a}+1\right)\left(\sqrt{a}-1-\sqrt{a}-1\right)}{a-1}\right)\)
= \(\frac{\left(a-1\right)^2}{4a}.\frac{2\sqrt{a}.\left(-2\right)}{a-1}\) = \(\frac{\left(a-1\right)^2}{4a}.\frac{-4\sqrt{a}}{a-1}\)
=\(\frac{\left(-4\sqrt{a}\right)\left(a-1\right)^2}{4a\left(a-1\right)}=\frac{\left(-4\sqrt{a}\right)\left(a-1\right)}{4a}\)
= \(\frac{-\sqrt{a}\left(a-1\right)}{a}=\frac{-\left(a\sqrt{a}-\sqrt{a}\right)}{a}=\frac{\sqrt{a}-a\sqrt{a}}{a}\)
