- Xin phép giải lại cho HOÀN CHỈNH
ĐKXĐ : \(\left\{{}\begin{matrix}a-\sqrt{a}\ne0\\a-1\ne0\\a\ge0\end{matrix}\right.\)=> \(\left\{{}\begin{matrix}\sqrt{a}\left(\sqrt{a}-1\right)\ne0\\\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)\ne0\\a\ge0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\sqrt{a}\ne0\\\sqrt{a}-1\ne0\\a\ge0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}a\ne0\\a\ne1\\a\ge0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)
Ta có : \(F=\left(\frac{1}{\sqrt{a}-1}+\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+1}+\frac{2}{a-1}\right)\)
=> \(F=\left(\frac{1}{\sqrt{a}-1}+\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{1}{\sqrt{a}+1}+\frac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
=> \(F=\left(\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\sqrt{a}-1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}+\frac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
=> \(F=\left(\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
=> \(F=\left(\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{1}{\sqrt{a}-1}\right)\)
=> \(F=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}=\frac{\sqrt{a}+1}{\sqrt{a}}\)
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