a, ĐKXĐ : \(\left\{{}\begin{matrix}a\ge0\\a+1\ne0\\\sqrt{a}-1\ne0\\a\sqrt{a}+\sqrt{a}-a-1\ne0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}a\ge0\\a\ne-1\\a\ne1\\\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)-\sqrt{x}\left(\sqrt{x}-1\right)\ne0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}a\ge0\\a-1\\a\ne1\\\left(\sqrt{x}-1\right)\left(x+1\right)\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}a\ge0\\a\ne-1\\a\ne1\end{matrix}\right.\)
- Ta có : \(A=\left(1+\frac{\sqrt{a}}{a+1}\right):\left(\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}-a-1}\right)\)
=> \(A=\left(1+\frac{\sqrt{a}}{a+1}\right):\left(\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{\left(\sqrt{a}\right)^3-1-a+\sqrt{a}}\right)\)
=> \(A=\left(1+\frac{\sqrt{a}}{a+1}\right):\left(\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)-\sqrt{a}\left(\sqrt{a}-1\right)}\right)\)
=> \(A=\left(1+\frac{\sqrt{a}}{a+1}\right):\left(\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1-\sqrt{a}\right)}\right)\)
=> \(A=\left(\frac{a+1}{a+1}+\frac{\sqrt{a}}{a+1}\right):\left(\frac{a+1}{\left(\sqrt{a}-1\right)\left(a+1\right)}-\frac{2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\right)\)
=> \(A=\left(\frac{a+1+\sqrt{a}}{a+1}\right):\left(\frac{a+1-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\right)\)
=> \(A=\left(\frac{a+1+\sqrt{a}}{a+1}\right).\left(\frac{\left(\sqrt{a}-1\right)\left(a+1\right)}{a+1-2\sqrt{a}}\right)\)
=> \(A=\frac{\left(\sqrt{a}-1\right)\left(a+1\right)\left(a+1+\sqrt{a}\right)}{\left(a+1-2\sqrt{a}\right)\left(a+1\right)}\)
=> \(A=\frac{\left(\sqrt{a}-1\right)\left(a+1+\sqrt{a}\right)}{\left(\sqrt{a}-1\right)^2}\)
=> \(A=\frac{a+1+\sqrt{a}}{\sqrt{a}-1}\)
@Phạm Lan Hương
@Nguyễn Ngọc Lộc