- ĐKXĐ : \(\left\{{}\begin{matrix}a\ge0\\\sqrt{a}-1\ne0\\\sqrt{a}+1\ne0\\2\sqrt{a}\ne0\end{matrix}\right.\)=> \(\left\{{}\begin{matrix}a\ne0\\a\ge0\\a\ne1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)
- Ta có phương trình : \(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)\left(\frac{a-\sqrt{a}}{\sqrt{a}+1}-\frac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
=\(\left(\frac{a}{2\sqrt{a}}-\frac{1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}+1}-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)^2}{a-1}-\frac{\sqrt{a}\left(\sqrt{a}+1\right)^2}{a-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)^2-\sqrt{a}\left(\sqrt{a}+1\right)^2}{a-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2\right)}{a-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(a-2\sqrt{a}+1-a-2\sqrt{a}-1\right)}{a-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(-4\sqrt{a}\right)}{a-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{-4a}{a-1}\right)\)= \(\frac{-4a\left(a-1\right)}{2\sqrt{a}\left(a-1\right)}\) = \(\frac{-4a}{2\sqrt{a}}\)
= \(\frac{-4\sqrt{a}\sqrt{a}}{2\sqrt{a}}\) = \(-2\sqrt{a}\)
